Q.If $x + iy = \dfrac{a + ib}{a - ib}$, prove that $x^{2} + y^{2} = 1$.

Showing the 12 most recent of 17 on this concept.

1. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.One of the values of $\left(\sqrt{3} - i\right)^{5/3}$ is (A) $2^{5/3} \text{cis} \dfrac{5\pi}{18}$ (B) $2^{5/3} \text{cis} \dfrac{19\pi}{18}$ (C) $2^{5/3} \text{cis} \dfrac{23\pi}{18}$ (D) $2^{5/3} \text{cis} \dfrac{17\pi}{18}$

   ›Reveal solutionSolution

   To find one of the values of a complex number raised to a fractional power, first convert the base complex number to its polar form. Then, apply De Moivre's theorem for roots, ensuring you account for the multiple possible arguments by adding $2k\pi$ to the original argument. The correct option is (B).

   When dealing with powers and roots of complex numbers, converting the number from Cartesian form ($x+iy$) to polar form ($r(\cos\theta + i\sin\theta)$ or $r \text{cis}\theta$) simplifies calculations significantly. This is because De Moivre's Theorem provides a direct way to compute such expressions.

   For a complex number $z = r \text{cis}\theta$, its $n$-th power is $z^n = r^n \text{cis}(n\theta)$. When $n$ is a fraction, say $p/q$, we are essentially looking for the $q$-th roots of $z^p$. A crucial point here is that the argument $\theta$ of a complex number is not unique; $\theta + 2k\pi$ (for any integer $k$) represents the same complex number. When taking roots, these different representations lead to distinct roots. Specifically, for $z^{p/q}$, the $q$ distinct values are given by:

   If $z = r \text{cis}\theta$, then $z^{p/q} = r^{p/q} \text{cis}\left(\frac{p(\theta + 2k\pi)}{q}\right)$, where $k = 0, 1, 2, \dots, q-1$.

   Let's apply this to the given problem.

   1. Convert the base complex number to polar form.

      The given complex number is $z = \sqrt{3} - i$.

      • Modulus ($r$): The modulus is the distance from the origin to the point $(\sqrt{3}, -1)$ in the complex plane. $r = |z| = \sqrt{(\text{Re}(z))^2 + (\text{Im}(z))^2} = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2$.
      • Argument ($\theta$): The argument is the angle the line segment from the origin to the point makes with the positive real axis. We have $\cos\theta = \frac{\text{Re}(z)}{r} = \frac{\sqrt{3}}{2}$ and $\sin\theta = \frac{\text{Im}(z)}{r} = \frac{-1}{2}$. Since $\cos\theta > 0$ and $\sin\theta < 0$, the complex number lies in the fourth quadrant. The reference angle (acute angle with the real axis) is $\alpha = \arctan\left(\left|\frac{-1}{\sqrt{3}}\right|\right) = \arctan\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$. In the fourth quadrant, the principal argument is $\theta = -\frac{\pi}{6}$. So, the complex number in polar form is $z = 2 \text{cis}\left(-\frac{\pi}{6}\right)$.

   2. Apply De Moivre's Theorem for fractional powers.

      We need to find the values of $z^{5/3}$. Here, $r=2$, $p=5$, $q=3$, and $\theta = -\frac{\pi}{6}$.

      Using the formula for fractional powers:

      $\left(\sqrt{3} - i\right)^{5/3} = \left(2 \text{cis}\left(-\frac{\pi}{6}\right)\right)^{5/3} = 2^{5/3} \text{cis}\left(\frac{5\left(-\frac{\pi}{6} + 2k\pi\right)}{3}\right)$, for $k=0, 1, 2$.

   3. Calculate the distinct values for $k=0, 1, 2$.

      We need to find one of the values, so we will calculate the arguments for $k=0, 1, 2$ and check which one matches the given options.

      • For $k=0$: Argument $= \frac{5\left(-\frac{\pi}{6} + 2(0)\pi\right)}{3} = \frac{5\left(-\frac{\pi}{6}\right)}{3} = -\frac{5\pi}{18}$. To express this in the standard range $[0, 2\pi)$, we add $2\pi$: $-\frac{5\pi}{18} + 2\pi = \frac{-5\pi + 36\pi}{18} = \frac{31\pi}{18}$. …

   🔒 Start your 14-day free trial to see the full solution →
2. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.If the curve represented by the locus of a point in the Argand plane corresponding to the complex number $z$ satisfying the relation $\operatorname{Re}\left(\frac{z+4}{2z-5}\right) + \operatorname{Re}\left(\frac{\overline{z}+4}{2\overline{z}-5}\right) = 4$ cuts the X-axis at two points A and B, then the sum of the abscissae of those two points is (A) $0$ (B) $\frac{5}{4}$ (C) $\frac{43}{6}$ (D) $\frac{53}{7}$

   ›Reveal solutionSolution

   The given condition simplifies to the equation of a circle in the complex plane; its intersection with the real axis gives two points whose abscissae sum to the x‑coordinate of the centre multiplied by 2, which yields the answer 5/4.

   We start with the relation

   $\operatorname{Re}\left(\frac{z+4}{2z-5}\right) + \operatorname{Re}\left(\frac{\overline{z}+4}{2\overline{z}-5}\right) = 4.$

   Concept & Intuition

   For any complex number $w$, $\operatorname{Re}(w) = \frac{w + \overline{w}}{2}$.

   Here the second term is exactly the conjugate of the first term (since conjugating the fraction $\frac{z+4}{2z-5}$ gives $\frac{\overline{z}+4}{2\overline{z}-5}$).

   Thus the sum of the two real parts is simply $2 \operatorname{Re}\left(\frac{z+4}{2z-5}\right)$.

   So the equation becomes

   $2 \operatorname{Re}\left(\frac{z+4}{2z-5}\right) = 4 \quad\Longrightarrow\quad \operatorname{Re}\left(\frac{z+4}{2z-5}\right) = 2.$

   Now set $z = x + iy$ (with $x,y$ real). The condition $\operatorname{Re}(w)=2$ means the real part of that fraction equals 2. This will produce an equation in $x$ and $y$ that turns out to be a circle. The problem asks where this curve cuts the X‑axis ($y=0$); those intersection points have abscissae whose sum we need.

   ---

   Step‑by‑step solution

   1. Write the fraction in terms of $x,y$

   $\frac{z+4}{2z-5} = \frac{(x+4)+iy}{(2x-5)+i(2y)}.$

   2. Rationalise to extract the real part Multiply numerator and denominator by the conjugate of the denominator:

   $\frac{(x+4+iy)((2x-5)-i(2y))}{(2x-5)^2 + (2y)^2}.$

   The real part of this expression is the real part of the numerator divided by the (real) denominator. So compute the numerator’s real part:

   $\text{Re} = (x+4)(2x-5) + (y)(2y) = (x+4)(2x-5) + 2y^2.$

   (The cross terms $i(\dots)$ are imaginary and vanish when taking the real part.)

   3. Set the real part equal to 2

   $\frac{(x+4)(2x-5) + 2y^2}{(2x-5)^2 + 4y^2} = 2.$

   4. Clear the denominator (note: denominator is positive except at the pole $z=5/2$, which is not on the locus)

   $(x+4)(2x-5) + 2y^2 = 2\bigl[(2x-5)^2 + 4y^2\bigr].$

   5. Expand both sides Left:

   $(x+4)(2x-5) = 2x^2 -5x +8x -20 = 2x^2 +3x -20,$

   so left side = $2x^2 + 3x -20 + 2y^2$.

   Right:

   $2\bigl[(2x-5)^2 + 4y^2\bigr] = 2\bigl[4x^2 -20x +25 + 4y^2\bigr] = 8x^2 -40x +50 + 8y^2.$

   6. Bring all terms to one side

   $2x^2 + 3x -20 + 2y^2 - 8x^2 + 40x -50 - 8y^2 = 0,$

   $$$$

   🔒 Start your 14-day free trial to see the full solution →
3. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.$\left( \sqrt{3} + i \right)^{10} + \left( \sqrt{3} - i \right)^{10} =$ (A) $1024 \sqrt{3}$ (B) $1024$ (C) $2048$ (D) $512 \sqrt{3}$

   ›Reveal solutionSolution

   Use De Moivre’s theorem to convert the complex numbers to polar form; the sum simplifies to $2 \cdot 2^{10} \cos(10 \cdot \pi/6) = 2048 \cdot \cos(5\pi/3) = 2048 \cdot \frac12 = 1024$, so the answer is (B).

   Concept & Intuition

   When you see powers of complex conjugates like $(\sqrt{3}+i)^n$ and $(\sqrt{3}-i)^n$, their sum is twice the real part of either term. That’s because for any complex number $z$, $z^n + \overline{z}^n = 2 \operatorname{Re}(z^n)$. So the problem reduces to finding the real part of $(\sqrt{3}+i)^{10}$ — and the cleanest way is to write $\sqrt{3}+i$ in polar form and use De Moivre’s theorem.

   Step-by-step solution

   1. Convert $\sqrt{3}+i$ to polar form. The modulus is

   $r = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = 2.$

   The argument $\theta$ satisfies $\cos\theta = \frac{\sqrt{3}}{2}$ and $\sin\theta = \frac12$, so $\theta = \frac{\pi}{6}$ (or $30^\circ$).

   Hence

   $\sqrt{3}+i = 2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right).$

   2. Apply De Moivre’s theorem for the 10th power.

   $(\sqrt{3}+i)^{10} = 2^{10}\left(\cos\frac{10\pi}{6} + i\sin\frac{10\pi}{6}\right) = 1024\left(\cos\frac{5\pi}{3} + i\sin\frac{5\pi}{3}\right).$

   Since $\frac{5\pi}{3} = 300^\circ$, we have $\cos\frac{5\pi}{3} = \frac12$ and $\sin\frac{5\pi}{3} = -\frac{\sqrt{3}}{2}$.

   So

   $(\sqrt{3}+i)^{10} = 1024\left(\frac12 - i\frac{\sqrt{3}}{2}\right) = 512 - 512i\sqrt{3}.$

   3. Use the conjugate relationship. The conjugate of $\sqrt{3}+i$ is $\sqrt{3}-i$, so $$$$

   🔒 Start your 14-day free trial to see the full solution →
4. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.The length of the chord of the ellipse $\dfrac{x^2}{4} + y^2 = 1$ formed on the line $y = x + 1$ is (A) $2\sqrt{2}$ (B) $\dfrac{4}{5}\sqrt{2}$ (C) $4\sqrt{2}$ (D) $\dfrac{8}{5}\sqrt{2}$

   ›Reveal solutionSolution

   The chord length is found by solving the intersection of the ellipse and the line, then applying the distance formula. The correct length is $\frac{8}{5}\sqrt{2}$, which corresponds to option (D).

   We are given the ellipse $\frac{x^2}{4} + y^2 = 1$ and the line $y = x + 1$. The chord is the segment of the line that lies inside the ellipse. To find its length, we need the coordinates of the two intersection points, then compute the distance between them.

   Concept and intuition:

   The chord length between two intersection points of a line and a conic can be found without explicitly solving for each point individually. By substituting the line equation into the ellipse, we get a quadratic in $x$. The difference between the two $x$-coordinates (or $y$-coordinates) can be expressed using the quadratic's coefficients, and then the distance formula gives the chord length. This avoids messy radicals.

   Step-by-step solution:

   1. Substitute the line into the ellipse. The ellipse is $\frac{x^2}{4} + y^2 = 1$. Replace $y$ with $x+1$:

   $\frac{x^2}{4} + (x+1)^2 = 1.$

   2. Simplify to a quadratic in $x$. Expand $(x+1)^2 = x^2 + 2x + 1$. Then:

   $\frac{x^2}{4} + x^2 + 2x + 1 = 1.$

   Multiply through by 4:

   $x^2 + 4x^2 + 8x + 4 = 4.$

   Simplify:

   $5x^2 + 8x + 4 = 4 \quad \Rightarrow \quad 5x^2 + 8x = 0.$

   So:

   $x(5x + 8) = 0.$

   3. Find the $x$-coordinates of intersection. The solutions are:

   $x = 0 \quad \text{and} \quad x = -\frac{8}{5}.$

   4. Find the corresponding $y$-coordinates using $y = x+1$.

      For $x = 0$: $y = 1$.

      For $x = -\frac{8}{5}$: $y = -\frac{8}{5} + 1 = -\frac{3}{5}$.

      So the intersection points are $(0, 1)$ and $\left(-\frac{8}{5}, -\frac{3}{5}\right)$.

   5. Compute the chord length using the distance formula.

      $$$$

   🔒 Start your 14-day free trial to see the full solution →
5. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.$\left(\sqrt{3} + i\right)^{10} + \left(\sqrt{3} - i\right)^{10} =$ (A) 2048 (B) $1024\sqrt{3}$ (C) $512\sqrt{3}$ (D) 1024

   ›Reveal solutionSolution

   The sum simplifies using De Moivre’s theorem: both terms are complex conjugates, so their sum is twice the real part. Converting to polar form gives $2^{10} \cdot 2\cos(10\pi/6) = 1024 \cdot 2\cos(5\pi/3) = 2048 \cdot \frac12 = 1024$. The correct option is (D).

   Concept and intuition

   When you see a sum of a complex number and its conjugate raised to the same power, you can exploit symmetry. For any complex number $z$, $z^n + \overline{z}^n = 2\,\text{Re}(z^n)$. So instead of expanding two messy binomials, we just find the real part of one term and double it. The numbers $\sqrt{3} \pm i$ are perfectly set up for polar form because their modulus and argument are nice.

   Step-by-step

   1. Identify the conjugate pair

      Let $z = \sqrt{3} + i$. Then $\overline{z} = \sqrt{3} - i$. The expression is $z^{10} + \overline{z}^{10}$. Since $\overline{z}^{10} = \overline{z^{10}}$, the sum is $2\,\text{Re}(z^{10})$.

   2. Convert $z$ to polar form

      Modulus: $|z| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = 2$.

      Argument: $\tan\theta = \frac{1}{\sqrt{3}}$, so $\theta = \frac{\pi}{6}$ (since both coordinates are positive).

      Hence $z = 2\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right)$.

   3. Apply De Moivre’s theorem

      $z^{10} = 2^{10}\left(\cos\frac{10\pi}{6} + i\sin\frac{10\pi}{6}\right) = 1024\left(\cos\frac{5\pi}{3} + i\sin\frac{5\pi}{3}\right)$.

   4. Find the real part …

   🔒 Start your 14-day free trial to see the full solution →
6. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.The length of the chord of the ellipse $\dfrac{x^2}{4} + y^2 = 1$ formed on the line $y = x + 1$ is (A) $2\sqrt{2}$ (B) $\dfrac{4}{5}\sqrt{2}$ (C) $\dfrac{8}{5}\sqrt{2}$ (D) $4\sqrt{2}$

   ›Reveal solutionSolution

   The chord length is found by solving the ellipse and line simultaneously, then using the distance formula with the quadratic’s discriminant. The result is $\frac{8}{5}\sqrt{2}$, so the correct option is (C).

   We have the ellipse $\frac{x^2}{4} + y^2 = 1$ and the line $y = x + 1$. The chord is the segment of the line that lies inside the ellipse. To find its length, we find the intersection points and compute the distance between them.

   Why this works:

   For a line and a conic, substituting the line equation into the conic gives a quadratic in $x$. The roots are the $x$-coordinates of the intersection points. The chord length is then $\sqrt{1+m^2} \cdot |x_1 - x_2|$, where $m$ is the line’s slope. The difference $|x_1 - x_2|$ comes from the quadratic’s discriminant: $|x_1 - x_2| = \frac{\sqrt{\Delta}}{|a|}$.

   ---

   1. Substitute the line into the ellipse The ellipse: $\frac{x^2}{4} + y^2 = 1$. The line: $y = x + 1$. Substitute:

   $\frac{x^2}{4} + (x+1)^2 = 1$

   Multiply through by 4:

   $x^2 + 4(x+1)^2 = 4$

   Expand:

   $x^2 + 4(x^2 + 2x + 1) = 4 \quad\Rightarrow\quad x^2 + 4x^2 + 8x + 4 = 4$

   Simplify:

   $5x^2 + 8x + 4 = 4 \quad\Rightarrow\quad 5x^2 + 8x = 0$

   So:

   $x(5x + 8) = 0$

   The roots are $x = 0$ and $x = -\frac{8}{5}$.

   2. Find the corresponding $y$-coordinates

      Using $y = x + 1$:

      • For $x = 0$: $y = 1$ → point $A(0, 1)$.
      • For $x = -\frac{8}{5}$: $y = -\frac{8}{5} + 1 = -\frac{3}{5}$ → point $B\left(-\frac{8}{5}, -\frac{3}{5}\right)$.

   3. Compute the chord length

      Distance between $A$ and $B$:

      $$$$

   🔒 Start your 14-day free trial to see the full solution →
7. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
   Q.In a triangle ABC, $\dfrac{a(r_1 + r_2 r_3)}{r_1 - r + r_2 + r_3} =$ (A) $\sqrt{r_1 r_2 r_3}$ (B) $r_1 r_2 + r_2 r_3 + r_3 r_1$ (C) $2(R + r)$ (D) $2 + \dfrac{r}{2R}$

   ›Reveal solutionSolution

   Using $rr_1+r_2r_3=bc$ and $r_1+r_2+r_3-r=4R$, the expression becomes $\dfrac{abc}{4R}=\Delta$, and $\Delta=\sqrt{r\,r_1r_2r_3}$ — the square-root option, (A).

   The concept first

   Everything about the four "touching circles" of a triangle can be written with just $\Delta$ (area) and $s$ (semi-perimeter):

   $r=\frac{\Delta}{s},\qquad r_1=\frac{\Delta}{s-a},\qquad r_2=\frac{\Delta}{s-b},\qquad r_3=\frac{\Delta}{s-c},$

   together with Heron's formula in the form

   $\Delta^2=s(s-a)(s-b)(s-c).$

   Two consequences are worth memorising because they turn up constantly:

   • $r_1+r_2+r_3-r=4R$ (this is exactly the denominator here);
   • $r\,r_1r_2r_3=\Delta^2$, i.e. $\Delta = \sqrt{r\,r_1 r_2 r_3}$;
   • $abc = 4R\Delta$ (from the sine rule).

   Step-by-step

   1. Simplify $r r_1$.

   $r\,r_1=\frac{\Delta}{s}\cdot\frac{\Delta}{s-a}=\frac{\Delta^2}{s(s-a)}=\frac{s(s-a)(s-b)(s-c)}{s(s-a)}=(s-b)(s-c).$

   2. Simplify $r_2 r_3$.

   $r_2 r_3=\frac{\Delta^2}{(s-b)(s-c)}=\frac{s(s-a)(s-b)(s-c)}{(s-b)(s-c)}=s(s-a).$

   3. Add them — a lovely cancellation:

   $r r_1+r_2r_3=(s-b)(s-c)+s(s-a)$

   $=\left[s^2-s(b+c)+bc\right]+\left[s^2-sa\right]=2s^2-s(a+b+c)+bc .$

   Since $a+b+c=2s$,

   $=2s^2-2s^2+bc=\boxed{bc}.$

   4. The denominator is the standard identity

   $r_1+r_2+r_3-r=4R .$

   5. Put the pieces together: …

   🔒 Start your 14-day free trial to see the full solution →
8. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.The point P denotes the complex number $z = x + iy$ in the Argand plane. If $\dfrac{2z - i}{z - 2}$ is a purely real number, then the equation of the locus of P is (A) $2x^2 + 2y^2 - 4x - y = 0$ (B) $x + 4y - 2 = 0 \ \&\ (x, y) \neq (2, 0)$ (C) $x - 4y - 2 = 0 \ \&\ (x, y) \neq (2, 0)$ (D) $x^2 + y^2 - 4x - 2y = 0$

   ›Reveal solutionSolution

   The condition that $\frac{2z-i}{z-2}$ is purely real forces its imaginary part to be zero, which simplifies to a linear equation $x - 4y - 2 = 0$, excluding the point where the denominator vanishes; the correct option is (C).

   Concept & Intuition

   A complex number is purely real when its imaginary part equals zero. For a rational expression in $z$, we can multiply numerator and denominator by the conjugate of the denominator to separate real and imaginary parts. Alternatively, we can use the fact that a complex number $w$ is real iff $w = \overline{w}$. Both approaches lead to the same condition: the cross‑product of the numerator and denominator (in terms of $x$ and $y$) must vanish. The denominator cannot be zero, so we exclude the point $z = 2$.

   Step‑by‑step solution

   1. Set up the condition

      Let $w = \dfrac{2z - i}{z - 2}$. We require $w$ to be purely real, i.e. $\operatorname{Im}(w) = 0$.

      Write $z = x + iy$, so $2z - i = 2x + i(2y - 1)$ and $z - 2 = (x - 2) + iy$.

   2. Use the conjugate method

      A complex number is real iff it equals its own conjugate: $w = \overline{w}$.

      Hence

   $\frac{2z - i}{z - 2} = \overline{\left(\frac{2z - i}{z - 2}\right)} = \frac{2\overline{z} + i}{\overline{z} - 2}.$

   Cross‑multiply (valid because $z \neq 2$):

   $(2z - i)(\overline{z} - 2) = (z - 2)(2\overline{z} + i).$

   3. Expand both sides

      Left: $(2x + i(2y-1))((x-2) - iy)$.

      Right: $((x-2) + iy)(2x + i(1-2y))$.

      Instead of expanding fully, note that equality of these two expressions implies the imaginary parts cancel. A cleaner path: subtract the right side from the left side and simplify.

      Compute the difference:

   $(2z - i)(\overline{z} - 2) - (z - 2)(2\overline{z} + i) = 0.$

   Expand each product:

   • $(2z - i)(\overline{z} - 2) = 2z\overline{z} - 4z - i\overline{z} + 2i$.
   • $(z - 2)(2\overline{z} + i) = 2z\overline{z} + iz - 4\overline{z} - 2i$.

   Subtract:

   $(2z\overline{z} - 4z - i\overline{z} + 2i) - (2z\overline{z} + iz - 4\overline{z} - 2i) = -4z - i\overline{z} + 2i - iz + 4\overline{z} + 2i.$

   Simplify:

   $-4z - iz + 4\overline{z} - i\overline{z} + 4i = 0.$

   4. Factor and substitute $z = x+iy$, $\overline{z}=x-iy$ Group terms: $-(4+i)z + (4 - i)\overline{z} + 4i = 0$. Substitute:

   $-(4+i)(x+iy) + (4-i)(x-iy) + 4i = 0.$

   Expand:

   • $-(4+i)(x+iy) = -4x -4iy -ix + y$ (since $i \cdot iy = -y$).
   • $(4-i)(x-iy) = 4x -4iy -ix - y$ (since $-i \cdot -iy = -y$).

   Sum the two expansions:

   $(-4x + y) + (4x - y) = 0$ for the real parts;

   $(-4iy - ix) + (-4iy - ix) = -8iy - 2ix$ for the imaginary parts.

   So the sum becomes $0 + (-8iy - 2ix) + 4i = 0$.

   Divide by $i$ (or equate imaginary parts):

   $-8y - 2x + 4 = 0$ → $2x + 8y - 4 = 0$ → $x + 4y - 2 = 0$? Wait, check sign:

   Actually $-8y - 2x + 4 = 0$ → multiply by -1: $8y + 2x - 4 = 0$ → $2x + 8y = 4$ → $x + 4y = 2$. That is $x + 4y - 2 = 0$.

   But this is option (B), not (C). Did we make a sign error? Let’s verify carefully.

   5. Re‑check the algebra

      A safer method: For a fraction $\frac{a+ib}{c+id}$ to be real, we need $ad + bc = 0$ (the numerator times the conjugate denominator’s imaginary part).

      Here $a = 2x$, $b = 2y-1$, $c = x-2$, $d = y$.

      Condition: $a d + b c = 0$ → $(2x)(y) + (2y-1)(x-2) = 0$.

      Compute: $2xy + (2y-1)(x-2) = 2xy + 2y(x-2) -1(x-2) = 2xy + 2xy -4y -x +2 = 4xy -4y -x +2 = 0$.

      That gives $4xy - 4y - x + 2 = 0$. Factor? $4y(x-1) - (x-2) = 0$? Not obviously linear. This suggests the earlier conjugate method might have an error.

      Let’s redo the conjugate method with more care.

      Correct conjugate approach:

      $w = \overline{w}$ gives

   $\frac{2z - i}{z - 2} = \frac{2\overline{z} + i}{\overline{z} - 2}.$

   Cross‑multiply:

   $(2z - i)(\overline{z} - 2) = (z - 2)(2\overline{z} + i).$

   Expand left: $2z\overline{z} - 4z - i\overline{z} + 2i$.

   Expand right: $2z\overline{z} + iz - 4\overline{z} - 2i$.

   Subtract right from left:

   $(2z\overline{z} - 4z - i\overline{z} + 2i) - (2z\overline{z} + iz - 4\overline{z} - 2i) = -4z - i\overline{z} + 2i - iz + 4\overline{z} + 2i.$

   Combine: $(-4z - iz) + (4\overline{z} - i\overline{z}) + 4i = 0$.

   Factor: $-(4+i)z + (4 - i)\overline{z} + 4i = 0$.

   Now substitute $z = x+iy$, $\overline{z}=x-iy$:

   $-(4+i)(x+iy) + (4-i)(x-iy) + 4i = 0$.

   Compute $-(4+i)(x+iy) = -[4x + 4iy + ix + i^2 y] = -[4x + 4iy + ix - y] = -4x -4iy -ix + y$.

   Compute $(4-i)(x-iy) = 4x -4iy -ix + i^2 y = 4x -4iy -ix - y$.

   Sum: $(-4x + y + 4x - y) + (-4iy -ix -4iy -ix) + 4i = 0 + (-8iy -2ix) + 4i = 0$.

   So $-2ix -8iy + 4i = 0$. Divide by $i$: $-2x -8y + 4 = 0$ → multiply by -1: $2x + 8y - 4 = 0$ → $x + 4y - 2 = 0$. …

   🔒 Start your 14-day free trial to see the full solution →
9. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
   Q.If $\omega$ is the complex cube root of unity and

   [!FORMULA] $\left(\frac{a + b\omega + c\omega^2}{c + a\omega + b\omega^2}\right)^k + \left(\frac{a + b\omega + c\omega^2}{b + a\omega^2 + c\omega}\right)^l = 2,$

   then $2k + l$ is always (A) divisible by 2 (B) divisible by 6 (C) divisible by 3 (D) divisible by 5

   ›Reveal solutionSolution

   The key idea is that both fractions inside the parentheses are cube roots of unity (specifically $\omega$ or $\omega^2$), so raising them to powers $k$ and $l$ yields values that are also powers of $\omega$. The condition that their sum equals 2 forces both terms to equal 1, which implies $k$ and $l$ are multiples of 3, making $2k + l$ always divisible by 3.

   We start by recalling that $\omega$ is a primitive complex cube root of unity, so $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$. This means $\omega^2 = \overline{\omega}$ and $\omega \neq 1$.

   The expression involves two fractions. Let’s denote:

   $X = \frac{a + b\omega + c\omega^2}{c + a\omega + b\omega^2}, \quad Y = \frac{a + b\omega + c\omega^2}{b + a\omega^2 + c\omega}.$

   The given equation is $X^k + Y^l = 2$.

   The trick is to notice that $X$ and $Y$ are actually cube roots of unity (or their reciprocals). Why? Because the numerator and denominator are cyclic permutations of each other, and using $1 + \omega + \omega^2 = 0$, we can simplify.

   1. Simplify $X$. Multiply numerator and denominator by $\omega$ (or use the relation $\omega^2 = -1 - \omega$). A cleaner approach: note that if we replace $a, b, c$ by $a, b\omega, c\omega^2$, the denominator becomes $c + a\omega + b\omega^2$ times something? Actually, let’s do it directly.

      Consider the numerator $N = a + b\omega + c\omega^2$. Multiply it by $\omega$:

   $\omega N = a\omega + b\omega^2 + c\omega^3 = a\omega + b\omega^2 + c.$

   That’s exactly the denominator $D = c + a\omega + b\omega^2$. So $D = \omega N$. Hence

   $X = \frac{N}{D} = \frac{N}{\omega N} = \frac{1}{\omega} = \omega^2.$

   (Since $\omega^3 = 1$, $1/\omega = \omega^2$.)

   2. Simplify $Y$. Similarly, $Y = \frac{a + b\omega + c\omega^2}{b + a\omega^2 + c\omega}$. Multiply numerator by $\omega^2$:

   $\omega^2 N = a\omega^2 + b\omega^3 + c\omega^4 = a\omega^2 + b + c\omega.$

   That’s exactly the denominator $b + a\omega^2 + c\omega$. So denominator = $\omega^2 N$, and

   $Y = \frac{N}{\omega^2 N} = \frac{1}{\omega^2} = \omega.$

   Thus we have:

   $X = \omega^2, \quad Y = \omega.$

   Tip

   This simplification works because the coefficients are cyclic shifts of $(a, b, c)$ and the property $\omega^3 = 1$ makes multiplication by $\omega$ cycle them.

   3. Plug into the equation. The given condition becomes:

   $(\omega^2)^k + (\omega)^l = 2.$

   Since $\omega$ and $\omega^2$ are non-real complex numbers (unless $k$ or $l$ are multiples of 3), their powers are also cube roots of unity: $1, \omega, \omega^2$. …

   🔒 Start your 14-day free trial to see the full solution →
10. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
    Q.If $i = \sqrt{-1}$, then $(1+i)^{10} + (1-i)^{10} =$ (A) $-64$ (B) $64$ (C) $0$ (D) $64i$

    ›Reveal solutionSolution

    To evaluate the sum of powers of complex numbers, convert each complex number to its polar form and then apply De Moivre's Theorem. The sum simplifies to $\boxed{0}$.

    When dealing with powers of complex numbers, especially high powers, direct expansion using the binomial theorem can be very tedious and error-prone. A much more elegant and efficient approach is to convert the complex number into its polar form. This is because raising a complex number in polar form to a power becomes a simple multiplication of the angle and raising the modulus to that power, thanks to De Moivre's Theorem.

    Let's break down the calculation:

    1. Convert $(1+i)$ to polar form:

       A complex number $z = x+iy$ can be written in polar form as $z = r(\cos\theta + i\sin\theta)$, where $r = |z| = \sqrt{x^2+y^2}$ is the modulus and $\theta = \arg(z)$ is the argument.

       For $1+i$:

       The real part is $x=1$ and the imaginary part is $y=1$.

       The modulus is $r = \sqrt{1^2+1^2} = \sqrt{1+1} = \sqrt{2}$.

       The argument $\theta$ is found from $\tan\theta = \frac{y}{x} = \frac{1}{1} = 1$. Since $(1,1)$ lies in the first quadrant, $\theta = \frac{\pi}{4}$.

       So, $1+i = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)$.

    2. Convert $(1-i)$ to polar form:

       For $1-i$:

       The real part is $x=1$ and the imaginary part is $y=-1$.

       The modulus is $r = \sqrt{1^2+(-1)^2} = \sqrt{1+1} = \sqrt{2}$.

       The argument $\theta$ is found from $\tan\theta = \frac{y}{x} = \frac{-1}{1} = -1$. Since $(1,-1)$ lies in the fourth quadrant, $\theta = -\frac{\pi}{4}$ (or $\frac{7\pi}{4}$). Using $-\frac{\pi}{4}$ is generally more convenient for calculations.

       So, $1-i = \sqrt{2}\left(\cos\left(-\frac{\pi}{4}\right) + i\sin\left(-\frac{\pi}{4}\right)\right)$.

       Recall that $\cos(-\theta) = \cos\theta$ and $\sin(-\theta) = -\sin\theta$. Thus, $1-i = \sqrt{2}\left(\cos\frac{\pi}{4} - i\sin\frac{\pi}{4}\right)$.

    3. Apply De Moivre's Theorem to $(1+i)^{10}$:

       De Moivre's Theorem states that for any integer $n$, $(r(\cos\theta + i\sin\theta))^n = r^n(\cos(n\theta) + i\sin(n\theta))$.

       Using this theorem for $(1+i)^{10}$:

       $(1+i)^{10} = \left[\sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)\right]^{10}$

       $= (\sqrt{2})^{10}\left(\cos\left(10 \cdot \frac{\pi}{4}\right) + i\sin\left(10 \cdot \frac{\pi}{4}\right)\right)$

       $= 2^{10/2}\left(\cos\left(\frac{5\pi}{2}\right) + i\sin\left(\frac{5\pi}{2}\right)\right)$

       $= 2^5\left(\cos\left(2\pi + \frac{\pi}{2}\right) + i\sin\left(2\pi + \frac{\pi}{2}\right)\right)$

       Since $\cos(2\pi + \alpha) = \cos\alpha$ and $\sin(2\pi + \alpha) = \sin\alpha$:

       $= 32\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right)$

       We know $\cos\frac{\pi}{2} = 0$ and $\sin\frac{\pi}{2} = 1$.

       $= 32(0 + i \cdot 1) = 32i$.

    4. Apply De Moivre's Theorem to $(1-i)^{10}$:

       Using the polar form $1-i = \sqrt{2}\left(\cos\left(-\frac{\pi}{4}\right) + i\sin\left(-\frac{\pi}{4}\right)\right)$: …

    🔒 Start your 14-day free trial to see the full solution →
11. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
    Q.Let $z = x + iy$ be a point in the Argand plane. If the amplitude of $\left(\frac{z-3}{z+2i}\right)$ is $\frac{\pi}{2}$, then the locus of $z$ is (A) a circle (B) a straight line (C) a semicircular arc not containing the origin (D) a semicircular arc containing the origin

    ›Reveal solutionSolution

    The condition that the amplitude (argument) of $\frac{z-3}{z+2i}$ is $\frac{\pi}{2}$ means the complex number is purely imaginary and positive, which geometrically forces $z$ to lie on a circle with the segment joining $3$ and $-2i$ as a diameter — but only the arc that makes the argument exactly $\frac{\pi}{2}$, which is a semicircular arc not containing the origin.

    Concept & Intuition

    When a complex number has argument $\frac{\pi}{2}$, it lies on the positive imaginary axis. So $\frac{z-3}{z+2i}$ is a positive imaginary number. That means the vectors from $z$ to $3$ and from $z$ to $-2i$ are perpendicular, with $z-3$ leading $z+2i$ by $90^\circ$. In geometry, if two points $A$ and $B$ are fixed, the locus of $P$ such that $\angle APB = 90^\circ$ is a circle with $AB$ as diameter. Here $A=3$ (real axis) and $B=-2i$ (imaginary axis). But we must also respect the sign of the argument — it’s exactly $\frac{\pi}{2}$, not $-\frac{\pi}{2}$ — so only one semicircular arc qualifies.

    Step-by-step reasoning

    1. Translate the amplitude condition The amplitude (argument) of $\frac{z-3}{z+2i}$ is $\frac{\pi}{2}$. This means

    $\arg\left(\frac{z-3}{z+2i}\right) = \frac{\pi}{2}.$

    Since $\arg\left(\frac{u}{v}\right) = \arg(u) - \arg(v)$ (mod $2\pi$), we have

    $\arg(z-3) - \arg(z+2i) = \frac{\pi}{2}.$

    So the vector from $z$ to $3$ is rotated $90^\circ$ counterclockwise relative to the vector from $z$ to $-2i$.

    2. Interpret geometrically

       Let $A = 3$ (point $(3,0)$) and $B = -2i$ (point $(0,-2)$). Then $z-3$ is the vector from $A$ to $z$, and $z+2i$ is the vector from $B$ to $z$. The condition $\arg(z-3) - \arg(z+2i) = \frac{\pi}{2}$ means the angle at $z$ between the lines $zA$ and $zB$ is $90^\circ$, with $zA$ rotated counterclockwise from $zB$.

    3. Recall the circle theorem

       The locus of points $z$ such that $\angle AzB = 90^\circ$ is the circle with $AB$ as diameter (Thales’ theorem). The endpoints are $A(3,0)$ and $B(0,-2)$. The midpoint is

    $\left(\frac{3+0}{2}, \frac{0+(-2)}{2}\right) = \left(1.5, -1\right),$

    and the radius is half the distance $AB$:

    $\frac{1}{2}\sqrt{(3-0)^2 + (0+2)^2} = \frac{1}{2}\sqrt{9+4} = \frac{\sqrt{13}}{2}.$

    So the full circle equation is

    $(x-1.5)^2 + (y+1)^2 = \frac{13}{4}.$

    4. Restrict to the correct arc …

    🔒 Start your 14-day free trial to see the full solution →
12. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
    Q.If $1, \alpha_1, \alpha_2, \alpha_3, \ldots, \alpha_{n-1}$ are $n^{\text{th}}$ roots of unity then $\sum_{1 \leq i < j \leq n-1} \alpha_i \alpha_j =$ (A) $1$ (B) $0$ (C) $-1$ (D) $i$

    ›Reveal solutionSolution

    Using Vieta on $z^n-1=0$: the pairwise product of all $n$ roots is $0$, and separating the pairs containing the root $1$ gives $\displaystyle\sum_{1\le i<j\le n-1}\alpha_i\alpha_j = 1$ — option (A).

    The $n$ roots of unity $1,\alpha_1,\alpha_2,\dots,\alpha_{n-1}$ are the roots of

    $z^n-1=0.$

    Apply Vieta's formulas.

    1. Sum of all roots = $0$ (coefficient of $z^{n-1}$ is $0$):

    $1+\alpha_1+\alpha_2+\cdots+\alpha_{n-1}=0 \;\Rightarrow\; \alpha_1+\cdots+\alpha_{n-1}=-1.$

    2. Sum of pairwise products of all $n$ roots = $0$ (coefficient of $z^{n-2}$ is $0$). Split these pairs into those that include the root $1$ and those that do not: …

    🔒 Start your 14-day free trial to see the full solution →