Q.The real value of $\alpha$ for which the expression $\dfrac{1-i\sin\alpha}{1+2i\sin\alpha}$ is purely real is:
(A) $(n+1)\dfrac{\pi}{2}$
(B) $(2n+1)\dfrac{\pi}{2}$
(C) $n\pi$
(D) None of these, where $n\in\mathbf{N}$

Showing the 12 most recent of 13 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.If $\sqrt{-4x+2i\sqrt{x^4+2x^2+9}} = \pm (a+ib)$, then $a^2+b^2-6=$ (A) $x^4$ (B) $2x^2$ (C) $4x$ (D) $x^4+2x^2$

   ›Reveal solutionSolution

   The problem asks for the value of $a^2+b^2-6$ given a complex number equation. By squaring both sides and using the property that the magnitude of a complex number squared is the sum of the squares of its real and imaginary parts, we find $a^2+b^2 = 2x^2+6$. The final result is $\boxed{2x^2}$.

   When dealing with square roots of complex numbers, especially when the result is given in the form $\pm (a+ib)$, a powerful technique is to use the magnitude property of complex numbers. If $z$ is a complex number and $w = \sqrt{z}$, then $z = w^2$. A key property of magnitudes is that for any complex number $w$, $|w^2| = |w|^2$. This means if $w = a+ib$, then $|w|^2 = |a+ib|^2 = a^2+b^2$. This relationship directly gives us $a^2+b^2$, which is often what is required in such problems.

   Let's apply this concept to the given problem.

   1. Identify the complex number and its square root:

      Let the complex number inside the square root be $Z$.

      So, $Z = -4x+2i\sqrt{x^4+2x^2+9}$.

      We are given that $\sqrt{Z} = \pm (a+ib)$.

   2. Square both sides of the given equation:

      Squaring both sides eliminates the square root on the left and the $\pm$ sign on the right:

      $Z = (\pm (a+ib))^2$

      $Z = (a+ib)^2$

   3. Apply the magnitude property:

      We know that for any complex number $w$, $|w^2| = |w|^2$.

      Here, $Z = (a+ib)^2$. Taking the magnitude of both sides:

      $|Z| = |(a+ib)^2|$

      $|Z| = |a+ib|^2$

      Since $|a+ib| = \sqrt{a^2+b^2}$, then $|a+ib|^2 = (\sqrt{a^2+b^2})^2 = a^2+b^2$.

      Therefore, we have:

      $|Z| = a^2+b^2$

   4. Calculate the magnitude of $Z$:

      The complex number $Z$ is given as $Z = -4x+2i\sqrt{x^4+2x^2+9}$.

      For a complex number $X+iY$, its magnitude is $|X+iY| = \sqrt{X^2+Y^2}$.

      Here, $X = -4x$ and $Y = 2\sqrt{x^4+2x^2+9}$.

      $|Z| = \sqrt{(-4x)^2 + (2\sqrt{x^4+2x^2+9})^2}$

      $|Z| = \sqrt{16x^2 + 4(x^4+2x^2+9)}$

      $|Z| = \sqrt{16x^2 + 4x^4+8x^2+36}$ …

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2. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.If the inclination of a straight line $x-y+1=0$ with another straight line L is $30^\circ$ and m is the slope of the line L, then $m^2+1=$ (A) $4m$ (B) $2m$ (C) $-2m$ (D) $-4m$

   ›Reveal solutionSolution

   The key idea is to use the angle-between-lines formula with the given inclination of $30^\circ$ and the slope of the given line. The result is $m^2 + 1 = -4m$, so the correct option is (D).

   The problem gives you a line $x - y + 1 = 0$ and says its inclination with another line L is $30^\circ$. Inclination here means the acute angle between the two lines. You are asked to find a relation involving the slope $m$ of line L.

   The concept you need is the formula for the angle $\theta$ between two lines with slopes $m_1$ and $m_2$:

   $\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|$

   This formula comes from the difference of their angles with the x-axis. The absolute value ensures we get the acute angle (or the smaller angle) between them.

   Let’s work through it step by step.

   1. Find the slope of the given line.

      The line is $x - y + 1 = 0$. Rewrite it as $y = x + 1$. So its slope is $m_1 = 1$.

   2. Set up the angle formula.

      The angle between the lines is $30^\circ$, so $\tan 30^\circ = \frac{1}{\sqrt{3}}$.

      Let the slope of L be $m$. Then:

   $\frac{1}{\sqrt{3}} = \left| \frac{m - 1}{1 + m \cdot 1} \right| = \left| \frac{m - 1}{1 + m} \right|$

   3. Remove the absolute value — two cases. The absolute value gives two possibilities:

   $\frac{m - 1}{1 + m} = \pm \frac{1}{\sqrt{3}}$

   Watch out

   A common mistake is to forget the $\pm$ sign. The absolute value means both positive and negative cases are possible, and each leads to a different slope. Both are valid mathematically, but the relation $m^2+1$ might be the same for both — we must check.

   4. Solve the positive case.

   $\frac{m - 1}{m + 1} = \frac{1}{\sqrt{3}}$

   Cross-multiply: $\sqrt{3}(m - 1) = m + 1$

   $\sqrt{3}m - \sqrt{3} = m + 1$

   $(\sqrt{3} - 1)m = \sqrt{3} + 1$

   $m = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$

   Rationalize: multiply numerator and denominator by $\sqrt{3} + 1$:

   $m = \frac{(\sqrt{3} + 1)^2}{3 - 1} = \frac{3 + 2\sqrt{3} + 1}{2} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$

   5. Solve the negative case.

   $\frac{m - 1}{m + 1} = -\frac{1}{\sqrt{3}}$

   Cross-multiply: $\sqrt{3}(m - 1) = -(m + 1)$

   $\sqrt{3}m - \sqrt{3} = -m - 1$

   $(\sqrt{3} + 1)m = \sqrt{3} - 1$

   $m = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

   Rationalize: multiply numerator and denominator by $\sqrt{3} - 1$:

   $m = \frac{(\sqrt{3} - 1)^2}{3 - 1} = \frac{3 - 2\sqrt{3} + 1}{2} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$

   6. Now compute $m^2 + 1$ for each case. For $m = 2 + \sqrt{3}$:

   $m^2 = (2 + \sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3}$

   So $m^2 + 1 = 8 + 4\sqrt{3}$.

   For $m = 2 - \sqrt{3}$: …

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3. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.If $z = \dfrac{(1-i)^2}{(-\sqrt{3}+i)^3}$ then the principal amplitude of $z$ is (A) $-\dfrac{\pi}{2}$ (B) $\dfrac{\pi}{2}$ (C) $\pi$ (D) $-\dfrac{\pi}{3}$

   ›Reveal solutionSolution

   The principal amplitude of $z$ is found by simplifying the complex fraction using polar form and exponent rules, yielding $-\frac{\pi}{2}$.

   We are given

   $z = \frac{(1-i)^2}{(-\sqrt{3}+i)^3}.$

   We need the principal amplitude (the principal argument) of $z$, i.e., the angle $\theta$ in $(-\pi, \pi]$ such that $z = |z| e^{i\theta}$.

   ---

   Concept and Intuition

   When a complex number is expressed as a ratio of powers, the argument is simply the difference of the arguments of numerator and denominator (mod $2\pi$).

   Instead of expanding everything algebraically (which is messy), we convert each factor to polar form:

   • $1-i$ is a point in the fourth quadrant.
   • $-\sqrt{3}+i$ is in the second quadrant.

   Then we apply De Moivre’s theorem for powers, subtract angles, and reduce to the principal range.

   ---

   Step-by-step solution

   1. Find the polar form of $1-i$ Modulus: $|1-i| = \sqrt{1^2 + (-1)^2} = \sqrt{2}$. Argument: $\tan\theta = \frac{-1}{1} = -1$, and since the point $(1,-1)$ is in quadrant IV, $\theta = -\frac{\pi}{4}$. So

   $1-i = \sqrt{2} \, e^{-i\pi/4}.$

   2. Square it

   $(1-i)^2 = (\sqrt{2})^2 \, e^{-i\pi/2} = 2 \, e^{-i\pi/2}.$

   So the numerator has argument $-\frac{\pi}{2}$.

   3. Find the polar form of $-\sqrt{3}+i$ Modulus: $\sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3+1} = 2$. Argument: $\tan\phi = \frac{1}{-\sqrt{3}} = -\frac{1}{\sqrt{3}}$. The reference angle is $\frac{\pi}{6}$, but the point $(-\sqrt{3}, 1)$ is in quadrant II, so

   $\phi = \pi - \frac{\pi}{6} = \frac{5\pi}{6}.$

   Hence

   $-\sqrt{3}+i = 2 \, e^{i5\pi/6}.$

   4. Cube it

   $(-\sqrt{3}+i)^3 = 2^3 \, e^{i \cdot 3 \cdot \frac{5\pi}{6}} = 8 \, e^{i \cdot \frac{15\pi}{6}} = 8 \, e^{i \cdot \frac{5\pi}{2}}.$

   Reduce the argument modulo $2\pi$: $\frac{5\pi}{2} - 2\pi = \frac{\pi}{2}$.

   So

   $(-\sqrt{3}+i)^3 = 8 \, e^{i\pi/2}.$

   5. Divide

   $z = \frac{2 \, e^{-i\pi/2}}{8 \, e^{i\pi/2}} = \frac{1}{4} \, e^{-i\pi/2 - i\pi/2} = \frac{1}{4} \, e^{-i\pi}.$

   So the argument of $z$ is $-\pi$ (or $\pi$ modulo $2\pi$).

   6. Principal amplitude …

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4. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.If $1+2i$ is a root of the equation $x^4 - 3x^3 + 8x^2 - 7x + 5 = 0$, then sum of the squares of the other roots is (A) $\dfrac{8}{3}$ (B) $-4 - 4i$ (C) $2 + i$ (D) $0$

   ›Reveal solutionSolution

   Since the polynomial has real coefficients, $1-2i$ is also a root. The "other" three roots are $1-2i,\,p,\,q$, and the sum of their squares is $-4-4i$ — option (B).

   Concept

   For a polynomial with real coefficients, complex roots occur in conjugate pairs, so $1+2i$ forces $1-2i$ to be a root. The remaining two roots $p,q$ are found through Vieta's formulas.

   Step-by-step solution

   Let the roots be $r_1=1+2i,\ r_2=1-2i,\ p,\ q$ for $x^4-3x^3+8x^2-7x+5=0$.

   1. Sum of roots $=3$: with $r_1+r_2=2$, we get $p+q=1$.
   2. Sum of pairwise products $=8$:

   $r_1r_2+(r_1+r_2)(p+q)+pq=8.$

   With $r_1r_2=(1+2i)(1-2i)=5$ and $r_1+r_2=2,\ p+q=1$:

   $5+2(1)+pq=8\Rightarrow pq=1.$

   3. Squares of the unknown pair: …

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5. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.$\lim_{n \to \infty} \left[ \left( 1 + \frac{1}{n^2} \right) \left( 1 + \frac{4}{n^2} \right) \left( 1 + \frac{9}{n^2} \right) \ldots

   (2) \right]^{\frac{1}{n}} =$(A)$16e^{-1}$(B)$2e^{\left(\frac{\pi - 4}{2}\right)}$(C)$2 \log 2 - 1$(D)$2 + e^{\left(\frac{\pi - 4}{2}\right)}$

   ›Reveal solutionSolution

   The limit is a Riemann sum disguised as a product; taking logs converts it into an integral of $\log(1+x^2)$ from $0$ to $1$, yielding $2\log 2 - 2 + \frac{\pi}{2}$, which exponentiates to $2e^{(\pi-4)/2}$, matching option (B).

   We are asked to evaluate

   $L = \lim_{n \to \infty} \left[ \prod_{k=1}^{n} \left(1 + \frac{k^2}{n^2}\right) \right]^{1/n}.$

   The product runs from $k=1$ to $n$ (the problem writes “(2)” but that’s a misprint; the pattern is clear).

   The key idea: a product raised to $1/n$ suggests taking logs to turn it into an average of logs, which for large $n$ becomes an integral.

   ---

   1. Take the natural logarithm Let

   $\log L = \lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^{n} \log\left(1 + \frac{k^2}{n^2}\right).$

   2. Recognize a Riemann sum The sum $\frac{1}{n}\sum_{k=1}^{n} f(k/n)$ with $f(x)=\log(1+x^2)$ converges to $\int_0^1 \log(1+x^2)\,dx$ as $n\to\infty$. So

   $\log L = \int_0^1 \log(1+x^2)\,dx.$

   3. Evaluate the integral Integrate by parts: let $u = \log(1+x^2)$, $dv = dx$. Then $du = \frac{2x}{1+x^2}dx$, $v = x$.

   $\int_0^1 \log(1+x^2)\,dx = \left[ x\log(1+x^2) \right]_0^1 - \int_0^1 \frac{2x^2}{1+x^2}\,dx.$

   The boundary term: $1\cdot\log(2) - 0 = \log 2$.

   For the remaining integral, rewrite:

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

   Hence

   $\int_0^1 \frac{2x^2}{1+x^2}\,dx = \int_0^1 \left(2 - \frac{2}{1+x^2}\right)dx = \left[2x - 2\arctan x\right]_0^1 = 2 - 2\cdot\frac{\pi}{4} = 2 - \frac{\pi}{2}.$

   4. Combine results $$$$

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6. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.If $z = x + iy$ and if the point P represents $z$ in the Argand plane, then the locus of $z$ satisfying the equation $|z-1| + |z+i| = 2$ is (A) $15x^2 - 2xy + 15y^2 - 16x + 16y - 48 = 0$ (B) $3x^2 + 2xy + 3y^2 - 4x - 4y = 0$ (C) $3x^2 - 2xy + 3y^2 - 4x + 4y = 0$ (D) $15x^2 + 2xy + 15y^2 + 16x - 16y - 48 = 0$

   ›Reveal solutionSolution

   The equation $|z-1| + |z+i| = 2$ defines an ellipse with foci at $(1,0)$ and $(0,-1)$ and constant sum $2$. Since the distance between foci equals $2$, the ellipse degenerates into the line segment joining them. The correct locus is given by option (C).

   The key insight is recognizing the geometric meaning of $|z - z_1| + |z - z_2| = \text{constant}$. This is the definition of an ellipse: the set of points whose sum of distances to two fixed points (foci) is constant. However, a degenerate case occurs when the constant sum equals the distance between the foci — then the ellipse collapses to the line segment joining them. Here, the foci are $1$ (point $(1,0)$) and $-i$ (point $(0,-1)$). Let’s check if that’s what happens.

   1. Identify the foci and the constant sum.

      The equation is $|z-1| + |z+i| = 2$.

      Focus $F_1 = 1$ corresponds to $(1,0)$.

      Focus $F_2 = -i$ corresponds to $(0,-1)$.

      The constant sum is $2a = 2$, so $a = 1$.

   2. Compute the distance between the foci.

      Distance $2c = |1 - (-i)| = |1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2}$.

      So $c = \frac{\sqrt{2}}{2}$.

   3. Compare $a$ and $c$.

      For a non-degenerate ellipse, we need $a > c$. Here $a = 1$ and $c \approx 0.707$, so $a > c$ — it seems like a proper ellipse. But wait: the constant sum is $2$, and the distance between foci is $\sqrt{2} \approx 1.414$. Since $2 > 1.414$, this is indeed a genuine ellipse, not a degenerate one. So we must find its Cartesian equation.

   4. Set up the equation in coordinates.

      Let $z = x + iy$. Then

      $|z-1| = \sqrt{(x-1)^2 + y^2}$,

      $|z+i| = \sqrt{x^2 + (y+1)^2}$.

      The equation is

   $\sqrt{(x-1)^2 + y^2} + \sqrt{x^2 + (y+1)^2} = 2.$

   5. Square both sides (carefully). Let $S = \sqrt{(x-1)^2 + y^2} + \sqrt{x^2 + (y+1)^2} = 2$. Square:

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

   Simplify the non-radical part:

   $(x-1)^2 + y^2 = x^2 - 2x + 1 + y^2$

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

   Sum = $2x^2 + 2y^2 - 2x + 2y + 2$.

   So we have:

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

   Subtract 2 from both sides:

   $2x^2 + 2y^2 - 2x + 2y + 2\sqrt{((x-1)^2 + y^2)(x^2 + (y+1)^2)} = 2.$

   Divide by 2:

   $x^2 + y^2 - x + y + \sqrt{((x-1)^2 + y^2)(x^2 + (y+1)^2)} = 1.$

   Isolate the radical:

   $\sqrt{((x-1)^2 + y^2)(x^2 + (y+1)^2)} = 1 - (x^2 + y^2 - x + y).$

   6. Square again. Square both sides:

   $((x-1)^2 + y^2)(x^2 + (y+1)^2) = \left(1 - (x^2 + y^2 - x + y)\right)^2.$

   Expand the left side:

   $(x^2 - 2x + 1 + y^2)(x^2 + y^2 + 2y + 1)$.

   Let $A = x^2 + y^2$. Then left side = $(A - 2x + 1)(A + 2y + 1)$.

   Expand:

   $= (A+1 - 2x)(A+1 + 2y)$

   $= (A+1)^2 + 2y(A+1) - 2x(A+1) - 4xy$

   $= (A+1)^2 + 2(A+1)(y - x) - 4xy$.

   Right side: $1 - (A - x + y) = 1 - A + x - y$.

   Square: $(1 - A + x - y)^2$.

   7. Set them equal and simplify.

   $(A+1)^2 + 2(A+1)(y - x) - 4xy = (1 - A + x - y)^2.$

   Notice that $1 - A + x - y = -(A - 1 - x + y)$. It’s easier to expand both sides directly in terms of $x,y$.

   Expand left:

   $(x^2+y^2+1)^2 + 2(x^2+y^2+1)(y-x) - 4xy$.

   Expand right:

   $(1 - x^2 - y^2 + x - y)^2$.

   Let’s compute step by step. Write $L = x^2+y^2$. Then left = $(L+1)^2 + 2(L+1)(y-x) - 4xy$.

   Right = $(1 - L + x - y)^2 = ( (1-L) + (x-y) )^2 = (1-L)^2 + 2(1-L)(x-y) + (x-y)^2$.

   Now equate:

   $(L+1)^2 + 2(L+1)(y-x) - 4xy = (1-L)^2 + 2(1-L)(x-y) + (x-y)^2.$

   Note $(L+1)^2 = L^2 + 2L + 1$, $(1-L)^2 = L^2 - 2L + 1$. …

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7. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
   Q.If the imaginary part of $\dfrac{2z+1}{iz+1}$ is $-2$, then the locus of the point representing $z$ in the Argand plane is (A) a circle (B) a straight line (C) a parabola (D) an ellipse

   ›Reveal solutionSolution

   The condition that the imaginary part of a complex expression is constant forces a linear relation between $x$ and $y$, so the locus is a straight line.

   The key idea is simple: when a complex expression is given and we are told something about its real or imaginary part, we write $z = x + iy$, simplify the expression into the form $A + iB$, and then set $B = -2$. That equation in $x$ and $y$ is the locus.

   Let’s do it step by step.

   1. Set up $z$ and substitute. Let $z = x + iy$, where $x, y \in \mathbb{R}$. Then

   $2z + 1 = 2(x + iy) + 1 = (2x + 1) + i(2y)$

   and

   $iz + 1 = i(x + iy) + 1 = ix + i^2 y + 1 = 1 - y + ix.$

   2. Form the fraction. We need

   $w = \frac{2z+1}{iz+1} = \frac{(2x+1) + i(2y)}{(1-y) + ix}.$

   3. Rationalise — multiply numerator and denominator by the conjugate of the denominator. The conjugate of $(1-y) + ix$ is $(1-y) - ix$. So

   $w = \frac{[(2x+1) + i(2y)] \cdot [(1-y) - ix]}{[(1-y) + ix] \cdot [(1-y) - ix]}.$

   The denominator is a real number:

   $(1-y)^2 + x^2.$

   4. Expand the numerator. Multiply term by term:

   $(2x+1)(1-y) + (2x+1)(-ix) + i(2y)(1-y) + i(2y)(-ix).$

   Compute each piece:

   • $(2x+1)(1-y) = (2x+1) - (2x+1)y$
   • $(2x+1)(-ix) = -i x(2x+1)$
   • $i(2y)(1-y) = i(2y - 2y^2)$
   • $i(2y)(-ix) = -i^2 (2xy) = +2xy$ (since $i^2 = -1$).

   So the numerator becomes:

   $[(2x+1)(1-y) + 2xy] \;+\; i\big[-x(2x+1) + 2y - 2y^2\big].$

   Simplify the real part:

   $(2x+1)(1-y) + 2xy = (2x+1) - (2x+1)y + 2xy = (2x+1) - 2xy - y + 2xy = 2x + 1 - y.$

   So the real part is $2x + 1 - y$.

   The imaginary part (coefficient of $i$) is:

   $-x(2x+1) + 2y - 2y^2 = -2x^2 - x + 2y - 2y^2.$

   5. Write $w$ in the form $A + iB$.

   $w = \frac{(2x + 1 - y) + i(-2x^2 - x + 2y - 2y^2)}{(1-y)^2 + x^2}.$

   Hence the imaginary part of $w$ is

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

   6. Apply the given condition. We are told $B = -2$. So

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

   Multiply both sides by the denominator (which is positive for all real $x, y$, so no sign issues):

   $-2x^2 - x + 2y - 2y^2 = -2\big[(1-y)^2 + x^2\big].$

   7. Simplify the equation. Expand the right side: …

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8. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
   Q.If $z = x + iy$ and the point P in the Argand plane represents $z$, then the locus of $z$ satisfying the equation $|z-2| + |z-2i| = 4$ is (A) $4x^2 + 3xy + 4y^2 - 6x - 6y + 8 = 0$ (B) $3x^2 + 2xy + 3y^2 - 8x - 8y + 6 = 0$ (C) $3x^2 + 2xy + 3y^2 - 8x - 8y = 0$ (D) $4x^2 + 3xy + 4y^2 - 6x - 6y = 0$

   ›Reveal solutionSolution

   The equation $|z-2|+|z-2i|=4$ describes an ellipse with foci at $(2,0)$ and $(0,2)$, and the sum of distances $4$; after squaring and simplifying, the locus is $3x^2+2xy+3y^2-8x-8y=0$, which matches option (C).

   The key idea is that the sum of distances from a point to two fixed points is constant — that’s the definition of an ellipse. Here, the two fixed points (foci) are $2$ (on the real axis) and $2i$ (on the imaginary axis). The constant sum is $4$. So the locus is an ellipse, but because the foci are not aligned horizontally or vertically, the equation will have an $xy$ term. We just need to derive it algebraically.

   1. Set up the equation in terms of $x$ and $y$. Let $z = x + iy$. Then $|z-2| = \sqrt{(x-2)^2 + y^2}$ and $|z-2i| = \sqrt{x^2 + (y-2)^2}$. The given condition is

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

   2. Isolate one square root and square. Move the second square root to the right:

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

   Square both sides:

   $(x-2)^2 + y^2 = 16 - 8\sqrt{x^2 + (y-2)^2} + x^2 + (y-2)^2.$

   3. Simplify the squared terms. Expand:

   $x^2 - 4x + 4 + y^2 = 16 - 8\sqrt{x^2 + (y-2)^2} + x^2 + y^2 - 4y + 4.$

   Cancel $x^2 + y^2$ on both sides:

   $-4x + 4 = 16 - 8\sqrt{x^2 + (y-2)^2} - 4y + 4.$

   Simplify constants: $4$ cancels on both sides, leaving

   $-4x = 16 - 8\sqrt{x^2 + (y-2)^2} - 4y.$

   Rearranging:

   $8\sqrt{x^2 + (y-2)^2} = 16 - 4y + 4x.$

   Divide by $4$:

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

   4. Square again to eliminate the remaining square root. Square both sides:

   $4\left[x^2 + (y-2)^2\right] = (4 + x - y)^2.$

   Expand the left:

   $4x^2 + 4(y^2 - 4y + 4) = 4x^2 + 4y^2 - 16y + 16.$

   Expand the right:

   $(4 + x - y)^2 = (x - y + 4)^2 = (x-y)^2 + 8(x-y) + 16.$

   And $(x-y)^2 = x^2 - 2xy + y^2$, so

   $\text{Right} = x^2 - 2xy + y^2 + 8x - 8y + 16.$

   5. Bring all terms to one side.

   $4x^2 + 4y^2 - 16y + 16 = x^2 - 2xy + y^2 + 8x - 8y + 16.$

   Cancel $16$ on both sides:

   $4x^2 + 4y^2 - 16y = x^2 - 2xy + y^2 + 8x - 8y.$

   Move everything to the left:

   $4x^2 + 4y^2 - 16y - x^2 + 2xy - y^2 - 8x + 8y = 0.$

   Combine like terms:

   • $4x^2 - x^2 = 3x^2$ …

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9. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
   Q.The locus of $z$ such that $\left| \frac{z-i}{z+i} \right| = 2$, where $z = x + iy$, is (A) $3x^2 + 3y^2 + 10y + 3 = 0$ (B) $3x^2 - 3y^2 - 10y - 3 = 0$ (C) $3x^2 + 3y^2 + 10y - 3 = 0$ (D) $x^2 + y^2 - 5y + 3 = 0$

   ›Reveal solutionSolution

   The condition $\left| \frac{z-i}{z+i} \right| = 2$ describes a circle (Apollonius circle) because the ratio of distances from $z$ to two fixed points is constant. Substituting $z = x+iy$ and simplifying yields $3x^2 + 3y^2 + 10y + 3 = 0$, which matches option (A).

   The key idea is that the equation $\left| \frac{z-i}{z+i} \right| = 2$ means the distance from $z$ to $i$ is twice the distance from $z$ to $-i$. This is a classic Apollonius circle: the set of points whose distances to two fixed points have a constant ratio (not equal to 1) is a circle. Here, the fixed points are $i$ and $-i$ on the imaginary axis.

   ---

   1. Write the condition in distance form. Let $z = x + iy$. Then

   $|z - i| = \sqrt{x^2 + (y-1)^2}, \quad |z + i| = \sqrt{x^2 + (y+1)^2}.$

   The given equation becomes

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

   2. Square both sides to remove the square roots.

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

   3. Cross-multiply and expand.

   $x^2 + (y-1)^2 = 4\left[x^2 + (y+1)^2\right].$

   Expand:

   $x^2 + y^2 - 2y + 1 = 4x^2 + 4(y^2 + 2y + 1).$

   That is:

   $x^2 + y^2 - 2y + 1 = 4x^2 + 4y^2 + 8y + 4.$

   4. Bring all terms to one side.

   $0 = 4x^2 + 4y^2 + 8y + 4 - x^2 - y^2 + 2y - 1.$

   Simplify:

   $0 = (4x^2 - x^2) + (4y^2 - y^2) + (8y + 2y) + (4 - 1).$

   So:

   $$$$

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10. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
    Q.If $e^{it} = \cos t + i \sin t$ and $e^{-it} = \cos t - i \sin t$ then $\cosh(x + iy) - \cosh(x - iy) =$ (A) $2 \sinh x \sinh y$ (B) $2i \sinh x \cos y$ (C) $2 \cosh x \cos y$ (D) $2i \sinh x \sin y$

    ›Reveal solutionSolution

    Use the definition of $\cosh$ in terms of exponentials, substitute $z = x+iy$ and $\bar{z}=x-iy$, and simplify using Euler’s formula. The result is $2i \sinh x \sin y$, which matches option (D).

    The core idea here is that $\cosh$ is defined via exponentials, just like $\cos$ and $\sin$ are. When you see a combination like $\cosh(x+iy) - \cosh(x-iy)$, you’re really looking at the difference of two complex exponentials — and that’s a perfect setup for Euler’s formula to kick in.

    Let’s walk through it step by step.

    1. Recall the definition of $\cosh$ For any complex number $z$,

    $\cosh z = \frac{e^{z} + e^{-z}}{2}.$

    This is the same definition you use for real numbers — it extends naturally to complex arguments.

    2. Write the two terms Let $z = x + iy$ and $\bar{z} = x - iy$. Then

    $\cosh(x+iy) = \frac{e^{x+iy} + e^{-(x+iy)}}{2} = \frac{e^{x} e^{iy} + e^{-x} e^{-iy}}{2},$

    $\cosh(x-iy) = \frac{e^{x-iy} + e^{-(x-iy)}}{2} = \frac{e^{x} e^{-iy} + e^{-x} e^{iy}}{2}.$

    3. Subtract them

    $\cosh(x+iy) - \cosh(x-iy) = \frac{1}{2} \left[ (e^{x} e^{iy} + e^{-x} e^{-iy}) - (e^{x} e^{-iy} + e^{-x} e^{iy}) \right].$

    Group the terms:

    $= \frac{1}{2} \left[ e^{x}(e^{iy} - e^{-iy}) + e^{-x}(e^{-iy} - e^{iy}) \right].$

    Notice the second bracket is just the negative of the first:

    $e^{-x}(e^{-iy} - e^{iy}) = - e^{-x}(e^{iy} - e^{-iy}).$

    So

    $\cosh(x+iy) - \cosh(x-iy) = \frac{1}{2} (e^{iy} - e^{-iy}) (e^{x} - e^{-x}).$

    4. Use Euler’s formula From the given identities, $$e^{iy} - e^{-iy} = 2i \sin y, …$$

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11. TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
    Q.If $\cos \theta = -\dfrac{1}{\sqrt{2}}$ and $\tan \theta = 1$, then the general value of $\theta$ is (A) $2n\pi + \dfrac{\pi}{4}, n = 0, 1, 2, 3, \ldots$ (B) $(2n+1)\pi + \dfrac{\pi}{4}, n = 0, 1, 2, 3, \ldots$ (C) $n\pi + \dfrac{\pi}{4}, n = 0, 1, 2, 3, \ldots$ (D) $n\pi \pm \dfrac{\pi}{4}, n = 0, 1, 2, 3, \ldots$

    ›Reveal solutionSolution

    The conditions $\cos\theta = -1/\sqrt{2}$ and $\tan\theta = 1$ force $\theta$ to lie in the third quadrant, where both sine and cosine are negative. The only angle satisfying both is $\theta = 5\pi/4$ (or $225^\circ$), whose general form is $(2n+1)\pi + \pi/4$. The correct option is (B).

    The key here is that a single trigonometric equation often has multiple families of solutions, but when two conditions are given simultaneously, the overlap narrows down to a specific quadrant and a specific set.

    Let’s see why.

    1. Interpret the given values.

       $\cos\theta = -\dfrac{1}{\sqrt{2}}$ tells us the reference angle is $\pi/4$ (since $\cos(\pi/4) = 1/\sqrt{2}$), but the negative sign means $\theta$ is in the second or third quadrant.

       $\tan\theta = 1$ tells us the reference angle is also $\pi/4$ (since $\tan(\pi/4) = 1$), but the positive sign means $\theta$ is in the first or third quadrant (where $\tan$ is positive).

    2. Find the common quadrant.

       The only quadrant where both conditions hold is the third quadrant — because that’s where $\cos$ is negative and $\tan$ is positive. So $\theta$ must be an angle whose terminal side lies in the third quadrant and whose reference angle is $\pi/4$.

    3. Write the specific angle in $[0, 2\pi)$.

       In the third quadrant, the angle with reference $\pi/4$ is

    $\theta = \pi + \frac{\pi}{4} = \frac{5\pi}{4}.$

    Check: $\cos(5\pi/4) = -\frac{1}{\sqrt{2}}$ and $\tan(5\pi/4) = 1$. It works.

    4. Generalize to all solutions. Since trigonometric functions are periodic with period $2\pi$, every solution is obtained by adding integer multiples of $2\pi$ to this base angle: $\theta = 2n\pi + \frac{5\pi}{4}, \quad n \in \mathbb{Z}.$ …

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12. TG EAPCET 2021Set eng-2021-08-06-FN1 markMCQ
    Q.If $\begin{vmatrix} \alpha & \beta & \gamma \\ a & b & c \\ l & m & n \end{vmatrix} = (-1)^K \begin{vmatrix} m & n & l \\ b & c & a \\ \beta & \gamma & \alpha \end{vmatrix}$, then the least value of $K$ is (A) 2 (B) 3 (C) 4 (D) 5

    ›Reveal solutionSolution

    Reaching the second determinant needs one row interchange plus a cyclic column shift (two interchanges) — $3$ interchanges in all, so $(-1)^K$ needs least $K=3$, option (B).

    Setup. Start from

    $D=\begin{vmatrix}\alpha&\beta&\gamma\\a&b&c\\l&m&n\end{vmatrix}.$

    Step 1 — swap rows $R_1\leftrightarrow R_3$ (one interchange, factor $-1$):

    $\begin{vmatrix}l&m&n\\a&b&c\\\alpha&\beta&\gamma\end{vmatrix}.$

    Step 2 — cyclic left shift of columns $(C_1\,C_2\,C_3)\to(C_2\,C_3\,C_1)$. A $3$-column cyclic shift is achieved by two adjacent column interchanges, giving

    $\begin{vmatrix}m&n&l\\b&c&a\\\beta&\gamma&\alpha\end{vmatrix},$

    which is exactly the target determinant. …

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