Q.If $\alpha$ and $\beta$ are different complex numbers with $|\beta| = 1$, then find $\left|\dfrac{\beta - \alpha}{1 - \bar{\alpha}\beta}\right|$.

Showing the 12 most recent of 67 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.$\left[\dfrac{\sin\frac{2\pi}{9} + i\cos\frac{2\pi}{9} + 1}{\sin\frac{2\pi}{9} - i\cos\frac{2\pi}{9} + 1}\right]^3 =$ (A) $\dfrac{1}{2}(\sqrt{3} - i)$ (B) $-\dfrac{1}{2}(1 - i\sqrt{3})$ (C) $\dfrac{1}{2}(1 - i\sqrt{3})$ (D) $-\dfrac{1}{2}(\sqrt{3} - i)$

   ›Reveal solutionSolution

   Rewriting $\sin\theta+i\cos\theta$ as $e^{i(90^\circ-\theta)}$ collapses the whole fraction to a single exponential, making the cube trivial. The answer is $-\frac12(\sqrt3-i)$.

   Concept and Intuition

   A mixed expression like $\sin\theta+i\cos\theta$ is not directly $e^{i\theta}$, but swapping the roles of sine and cosine via the complementary-angle identities converts it into a clean exponential. Once both numerator and denominator are exponentials, ratios and powers become simple angle arithmetic.

   Step-by-Step Solution

   1. Let $\theta=\dfrac{2\pi}{9}=40^\circ$ and set $\phi=90^\circ-\theta=50^\circ$.
   2. Since $\sin\theta=\cos(90^\circ-\theta)=\cos\phi$ and $\cos\theta=\sin(90^\circ-\theta)=\sin\phi$:

   $\sin\theta+i\cos\theta=\cos\phi+i\sin\phi=e^{i\phi},\qquad \sin\theta-i\cos\theta=\cos\phi-i\sin\phi=e^{-i\phi}.$

   3. So the bracket becomes $\dfrac{e^{i\phi}+1}{e^{-i\phi}+1}$.
   4. Factor: $e^{i\phi}+1=e^{i\phi/2}\left(e^{i\phi/2}+e^{-i\phi/2}\right)=2\cos(\phi/2)e^{i\phi/2}$, and similarly $e^{-i\phi}+1=2\cos(\phi/2)e^{-i\phi/2}$. …

   🔒 Start your 14-day free trial to see the full solution →
2. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.If $(1 + \sin\alpha + i\cos\alpha)^8 = A(\cos\theta + i\sin\theta)$, then $A$ and $\theta$ are respectively (A) $2^8 \cos^8\left(\dfrac{\pi}{4} - \dfrac{\alpha}{2}\right), 2\alpha$ (B) $2^8 \cos^8\left(\dfrac{\pi}{4} - \dfrac{\alpha}{2}\right), -4\alpha$ (C) $2^8 \cos^8\left(\dfrac{\pi}{4} + \dfrac{\alpha}{2}\right), 2\alpha$ (D) $2^8 \cos^8\left(\dfrac{\pi}{4} + \dfrac{\alpha}{2}\right), -4\alpha$

   ›Reveal solutionSolution

   This tests converting a complex expression into modulus–argument (polar) form using the half-angle trick $1+\cos\varphi+i\sin\varphi = 2\cos(\varphi/2)e^{i\varphi/2}$, then applying De Moivre's theorem. The answer is $A=2^8\cos^8(\pi/4-\alpha/2)$, $\theta=-4\alpha$.

   Concept and Intuition

   Whenever you see an expression like $1+\cos\varphi + i\sin\varphi$, it's worth recognising the standard identity that factors it into $2\cos(\varphi/2)$ times a pure phase $e^{i\varphi/2}$ — this converts an awkward sum into a clean modulus-argument form that can be raised to a power instantly via De Moivre's theorem. Here $\sin\alpha$ and $\cos\alpha$ are first rewritten as $\cos$ and $\sin$ of the complementary angle so the identity applies directly.

   Step-by-Step Solution

   1. Use complementary-angle identities: $\sin\alpha=\cos\left(\dfrac{\pi}{2}-\alpha\right)$ and $\cos\alpha=\sin\left(\dfrac{\pi}{2}-\alpha\right)$. Let $\varphi = \dfrac{\pi}{2}-\alpha$. Then $1+\sin\alpha+i\cos\alpha = 1+\cos\varphi+i\sin\varphi$.
   2. Apply the half-angle factorization: $1+\cos\varphi = 2\cos^2(\varphi/2)$ and $\sin\varphi = 2\sin(\varphi/2)\cos(\varphi/2)$, so

   $1+\cos\varphi+i\sin\varphi = 2\cos(\varphi/2)\left[\cos(\varphi/2)+i\sin(\varphi/2)\right].$

   3. Substitute back $\varphi/2 = \dfrac{1}{2}\left(\dfrac{\pi}{2}-\alpha\right)=\dfrac{\pi}{4}-\dfrac{\alpha}{2}$. So $1+\sin\alpha+i\cos\alpha = 2\cos\left(\dfrac{\pi}{4}-\dfrac{\alpha}{2}\right)\left[\cos\left(\dfrac{\pi}{4}-\dfrac{\alpha}{2}\right)+i\sin\left(\dfrac{\pi}{4}-\dfrac{\alpha}{2}\right)\right]$ — this is modulus $r=2\cos\left(\dfrac{\pi}{4}-\dfrac{\alpha}{2}\right)$ with argument $\dfrac{\pi}{4}-\dfrac{\alpha}{2}$.
   4. Raise to the 8th power using De Moivre's theorem: $r^8\left[\cos\left(8\left(\dfrac{\pi}{4}-\dfrac{\alpha}{2}\right)\right)+i\sin\left(8\left(\dfrac{\pi}{4}-\dfrac{\alpha}{2}\right)\right)\right]$, where $r^8 = 2^8\cos^8\left(\dfrac{\pi}{4}-\dfrac{\alpha}{2}\right) = A$. …

   🔒 Start your 14-day free trial to see the full solution →
3. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.If $z_1 = 8 + 4i$, $z_2 = 6 + 4i$ and $Arg\left(\dfrac{z - z_1}{z - z_2}\right) = \dfrac{\pi}{4}$, then 'z' satisfies (A) $|z - 7 - 4i| = 1$ (B) $|z - 7 - 5i| = \sqrt{2}$ (C) $|z - 4i| = 8$ (D) $|z - 1 - 7i| = \sqrt{18}$

   ›Reveal solutionSolution

   The condition $\operatorname{Arg}\left(\frac{z - z_1}{z - z_2}\right) = \frac{\pi}{4}$ means the directed angle $\angle z_2 z z_1$ is $45^\circ$, so $z$ lies on an arc of a circle through $z_1$ and $z_2$. Using the geometry of the chord and the inscribed angle, we find the center and radius, leading to $|z - (7 + 5i)| = \sqrt{2}$, which matches option (B).

   ---

   The key insight: For complex numbers, $\operatorname{Arg}\left(\frac{z - z_1}{z - z_2}\right)$ is the directed angle $\angle z_2 z z_1$ — the angle at $z$ between the vectors to $z_1$ and $z_2$. When this angle is constant (and not $0$ or $\pi$), the point $z$ traces an arc of a circle passing through $z_1$ and $z_2$. This is the inscribed angle theorem in reverse: points that see a fixed chord under a fixed angle lie on a circular arc.

   Here the chord is $z_1 z_2$, and the fixed angle is $45^\circ$. So $z$ lies on a circle whose center is somewhere on the perpendicular bisector of $z_1 z_2$, and whose radius is determined by the chord length and the inscribed angle.

   ---

   1. Find the chord and its midpoint.

      $z_1 = 8 + 4i$, $z_2 = 6 + 4i$. Their difference is $2$, purely real, so the segment is horizontal of length $2$. The midpoint is $M = \frac{z_1 + z_2}{2} = 7 + 4i$.

   2. Relate the inscribed angle to the central angle.

      For a chord of length $d$, the inscribed angle $\theta$ subtended by the chord at a point on the circle is half the central angle subtended by the same chord. So the central angle $2\theta = 90^\circ$. That means the chord is seen from the center at a right angle.

   3. Find the distance from the midpoint to the center.

      If the chord subtends a central angle of $90^\circ$, then the chord and the two radii to its endpoints form an isosceles right triangle. The chord is the hypotenuse, so the distance from the midpoint to the center (the altitude in that triangle) is half the chord length:

   $\text{distance} = \frac{\text{chord length}}{2} = \frac{2}{2} = 1.$

   The center lies on the perpendicular bisector of the chord. Since the chord is horizontal, the perpendicular bisector is vertical through $M = 7 + 4i$. So the center is at $7 + 4i \pm 1$ in the imaginary direction — i.e., $7 + 5i$ or $7 + 3i$.

   4. Which center gives the correct argument? …

   🔒 Start your 14-day free trial to see the full solution →
4. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.If the point $P$ represents a complex number 'z' in the Argand diagram and $\dfrac{z-i}{z-1}$ is always purely imaginary, then the locus of 'P' is (A) Circle with centre $\left(\dfrac{1}{2}, \dfrac{1}{2}\right)$ and radius $\dfrac{1}{\sqrt{2}}$ (B) Circle with centre $\left(-\dfrac{1}{2}, -\dfrac{1}{2}\right)$ and radius $\dfrac{1}{\sqrt{2}}$ (C) Circle with centre $\left(\dfrac{1}{2}, \dfrac{1}{2}\right)$ and radius $\dfrac{1}{\sqrt{2}}$ except the points $(0,1)$ and $(1,0)$ (D) Circle with centre $\left(-\dfrac{1}{2}, -\dfrac{1}{2}\right)$ and radius $\dfrac{1}{2}$ except the point $(1,0)$

   ›Reveal solutionSolution

   The locus is the circle on diameter joining $(0,1)$ and $(1,0)$, minus those two end points.

   Concept and Intuition

   A ratio $\dfrac{z-a}{z-b}$ is purely imaginary when the argument between $(z-a)$ and $(z-b)$ is $\pm\tfrac{\pi}{2}$; geometrically $z$ sees the segment $ab$ at a right angle, so $z$ lies on the circle with $ab$ as diameter. Here $a=i=(0,1)$ and $b=1=(1,0)$.

   Step-by-Step Solution

   1. Set the real part of $\dfrac{z-i}{z-1}$ to zero (purely imaginary condition).
   2. This gives the right-angle/diameter circle through $(0,1)$ and $(1,0)$.
   3. Centre $=$ midpoint $=\left(\tfrac12,\tfrac12\right)$.
   4. Radius $=\tfrac12\sqrt{(1-0)^2+(0-1)^2}=\tfrac{\sqrt2}{2}=\tfrac{1}{\sqrt2}$.
   5. At $z=1$ the expression is undefined and at $z=i$ it is $0$ (real, not purely imaginary), so exclude $(1,0)$ and $(0,1)$.

   Common Mistakes

   • Choosing the same circle without deleting the two special points (option A), ignoring that the expression is undefined or real there. …

   🔒 Start your 14-day free trial to see the full solution →
5. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.The solutions of the equation $2\sqrt{2}x^4 = (\sqrt{3} - 1) + i(\sqrt{3} + 1)$ are (A) $x = \pm \operatorname{cis}\dfrac{3\pi}{38}, \pm \operatorname{cis}\dfrac{23\pi}{38}$ (B) $x = \pm \operatorname{cis}\dfrac{5\pi}{48}, \pm \operatorname{cis}\dfrac{29\pi}{48}$ (C) $x = \pm \operatorname{cis}\dfrac{7\pi}{48}, \pm \operatorname{cis}\dfrac{41\pi}{48}$ (D) $x = \pm \operatorname{cis}\dfrac{9\pi}{62}, \pm \operatorname{cis}\dfrac{27\pi}{62}$

   ›Reveal solutionSolution

   The key idea is to rewrite the complex number on the right in polar form, then take fourth roots by dividing the argument by 4 and adjusting for periodicity. The correct answer is option (B).

   We are solving

   $2\sqrt{2}\, x^4 = (\sqrt{3} - 1) + i(\sqrt{3} + 1).$

   The left side is a real multiple of $x^4$; the right side is a complex number. The natural approach: isolate $x^4$, then take fourth roots using De Moivre’s theorem. But first we must express the right-hand side in polar form $r(\cos\theta + i\sin\theta)$.

   ---

   1. Find the modulus of the right-hand side. Let $z = (\sqrt{3} - 1) + i(\sqrt{3} + 1)$. Its modulus is

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

   Compute:

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

   $(\sqrt{3} + 1)^2 = 3 + 2\sqrt{3} + 1 = 4 + 2\sqrt{3}.$

   Sum: $(4 - 2\sqrt{3}) + (4 + 2\sqrt{3}) = 8$.

   So $|z| = \sqrt{8} = 2\sqrt{2}$.

   Tip

   The modulus $2\sqrt{2}$ exactly matches the coefficient on the left. This is no coincidence — it suggests the equation simplifies nicely.

   2. Find the argument $\theta$ of $z$. We have

   $\tan\theta = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}.$

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

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

   So $\tan\theta = 2 + \sqrt{3}$.

   Recognize that $\tan 75^\circ = \tan(45^\circ + 30^\circ) = \frac{1 + 1/\sqrt{3}}{1 - 1/\sqrt{3}} = 2 + \sqrt{3}$.

   Thus $\theta = 75^\circ = \frac{5\pi}{12}$ (since both real and imaginary parts are positive, the angle is in the first quadrant).

   Hence

   $z = 2\sqrt{2} \left( \cos\frac{5\pi}{12} + i\sin\frac{5\pi}{12} \right).$

   3. Substitute into the equation.

   $2\sqrt{2}\, x^4 = 2\sqrt{2} \left( \cos\frac{5\pi}{12} + i\sin\frac{5\pi}{12} \right).$

   Cancel $2\sqrt{2}$ (nonzero):

   $x^4 = \cos\frac{5\pi}{12} + i\sin\frac{5\pi}{12} = \operatorname{cis}\frac{5\pi}{12}.$

   4. Take fourth roots. By De Moivre’s theorem, the fourth roots are

   $x = \operatorname{cis}\left( \frac{5\pi}{12} + 2k\pi \over 4 \right), \quad k = 0,1,2,3.$

   That is:

   $x = \operatorname{cis}\left( \frac{5\pi}{48} + \frac{k\pi}{2} \right).$

   For $k=0$: $\operatorname{cis}\frac{5\pi}{48}$. …

   🔒 Start your 14-day free trial to see the full solution →
6. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.Among the solutions of the equation $x^6 = 64$, the sum of all those solutions whose real part is negative is (A) $0$ (B) $-2$ (C) $-4$ (D) $-3$

   ›Reveal solutionSolution

   The six sixth-roots of $64$ are $2e^{i\pi k/3}$; the three with negative real part sum to $-4$.

   Find the roots. Since $64=2^6$, the solutions of $x^6=64$ are the sixth roots:

   $x_k = 2\,e^{i\pi k/3},\qquad k=0,1,2,3,4,5.$

   Listing them:

   $k$	root	real part
   $0$	$2$	$2$
   $1$	$1+i\sqrt3$	$1$
   $2$	$-1+i\sqrt3$	$-1$
   $3$	$-2$	$-2$
   $4$	$-1-i\sqrt3$	$-1$

   🔒 Start your 14-day free trial to see the full solution →
7. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.The modulus of the product of all the values of $(2+3i)^{(3/5)}$ is (A) $\sqrt{2197}$ (B) $\sqrt{2245}$ (C) $\sqrt{135}$ (D) $\sqrt{489}$

   ›Reveal solutionSolution

   The problem asks for the modulus of the product of all five distinct values of $(2+3i)^{3/5}$. The key idea: the modulus of a complex number raised to a rational power is independent of the argument choices, so the product of all fifth roots simplifies to a single real number. The final result is $\sqrt{2197}$, which corresponds to option (A).

   ---

   Concept and Intuition

   When we raise a complex number to a fractional power like $3/5$, we get multiple distinct values because the exponent corresponds to taking a 5th root and then cubing. Each value comes from a different choice of the argument (angle) of the original complex number. The modulus (distance from the origin) of each of these values is the same — only the angles differ. So the product of all five values will have a modulus equal to the product of their moduli. Since all moduli are equal, this is just $(\text{modulus of one value})^5$. But there's a deeper trick: the product of all 5th roots of a number is actually just that number itself (up to sign), because the roots are the solutions to $z^5 = w$, and their product equals $w$ (by Vieta). Here we have $(2+3i)^{3/5}$, so the product of all five values is exactly $(2+3i)^3$. Then we just need the modulus of that cube.

   ---

   Step-by-step solution

   1. Understand the expression We want all values of $(2+3i)^{3/5}$. Write $2+3i$ in polar form:

   $2+3i = r e^{i\theta}, \quad r = \sqrt{2^2 + 3^2} = \sqrt{13}, \quad \theta = \arctan(3/2).$

   Then

   $(2+3i)^{3/5} = \left( r e^{i\theta} \right)^{3/5} = r^{3/5} e^{i(3\theta/5 + 2\pi k \cdot 3/5)} \quad \text{for } k = 0,1,2,3,4.$

   The factor $2\pi k$ comes from adding multiples of $2\pi$ to the argument before taking the root. So there are five distinct values, each with the same modulus $r^{3/5} = (\sqrt{13})^{3/5} = 13^{3/10}$.

   2. Product of all five values Let the five values be $z_0, z_1, z_2, z_3, z_4$. Their product is:

   $\prod_{k=0}^{4} z_k = \prod_{k=0}^{4} \left( r^{3/5} e^{i(3\theta/5 + 6\pi k/5)} \right) = r^{3} e^{i\left(5 \cdot \frac{3\theta}{5} + \frac{6\pi}{5} \sum_{k=0}^{4} k \right)}.$

   The sum of $k$ from 0 to 4 is $10$, so the exponent becomes $i(3\theta + 12\pi)$. Since $e^{i12\pi} = 1$, we get:

   $$$$

   🔒 Start your 14-day free trial to see the full solution →
8. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.One of the values of the square root of $\left(\dfrac{1}{2} + \dfrac{\sqrt{3}\,i}{2}\right)$ (A) $\dfrac{\sqrt{3}}{2} - \dfrac{i}{2}$ (B) $\dfrac{1}{2} - \dfrac{\sqrt{3}\,i}{2}$ (C) $-\dfrac{\sqrt{3}}{2} + \dfrac{i}{2}$ (D) $-\dfrac{\sqrt{3}}{2} - \dfrac{i}{2}$

   ›Reveal solutionSolution

   Convert the complex number to polar form and halve both the modulus and the argument (plus the second root at argument $+\pi$) to get the two square roots; one of them matches option (D).

   Concept and Intuition

   To find square roots of a complex number, it's far easier to work in polar form: if $z = r(\cos\theta+i\sin\theta)$, then its square roots are $\sqrt{r}\left(\cos\frac{\theta+2k\pi}{2}+i\sin\frac{\theta+2k\pi}{2}\right)$ for $k=0,1$.

   Step-by-Step Solution

   1. Given $z = \dfrac{1}{2}+\dfrac{\sqrt3}{2}i$. Compute modulus: $r=\sqrt{(1/2)^2+(\sqrt3/2)^2}=\sqrt{1/4+3/4}=1$.
   2. Compute argument: $\theta = \tan^{-1}\left(\dfrac{\sqrt3/2}{1/2}\right)=\tan^{-1}(\sqrt3)=\dfrac{\pi}{3}$ (both real and imaginary parts positive, so it's in the first quadrant). So $z=\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}$.
   3. Square roots have modulus $\sqrt{1}=1$ and argument $\dfrac{\pi/3}{2}=\dfrac{\pi}{6}$, or $\dfrac{\pi}{6}+\pi = \dfrac{7\pi}{6}$.
   4. First root: $\cos\frac{\pi}{6}+i\sin\frac{\pi}{6} = \dfrac{\sqrt3}{2}+\dfrac{i}{2}$. …

   🔒 Start your 14-day free trial to see the full solution →
9. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.If $1, i \in C$, the set of complex numbers, then the value of $\sqrt{1} + \sqrt{i}$ cannot be (A) $\dfrac{\sqrt{2}+1+i}{\sqrt{2}}$ (B) $\dfrac{\sqrt{2}+1-i}{\sqrt{2}}$ (C) $\dfrac{\sqrt{2}-1-i}{\sqrt{2}}$ (D) $\dfrac{-\sqrt{2}-1-i}{\sqrt{2}}$

   ›Reveal solutionSolution

   The square root of a complex number is two-valued. √1 = ±1 and √i = ±(1+i)/√2. Combining every choice of √1 with every choice of √i gives four possible sums; the one printed value that does not appear among them is the answer — option (B).

   Every non-zero complex number has exactly two square roots, differing by a sign, so "the value of √1 + √i" must be checked over all four combinations before deciding which printed expression is impossible.

   1. Square roots of 1

   1 = e^{i·0}, so its square roots are e^{i·0} = 1 and e^{iπ} = -1. So √1 ∈ {1, -1}.

   2. Square roots of i

   i = e^{iπ/2}, so its square roots are e^{iπ/4} = (1+i)/√2 and its negative, -(1+i)/√2. So √i ∈ { (1+i)/√2, -(1+i)/√2 }.

   3. The four possible sums

   • 1 + (1+i)/√2 = (√2+1+i)/√2 — matches option (A)
   • 1 - (1+i)/√2 = (√2-1-i)/√2 — matches option (C)
   • -1 + (1+i)/√2 = (-√2+1+i)/√2 — not printed as any option
   • -1 - (1+i)/√2 = (-√2-1-i)/√2 — matches option (D)

   4. Compare with the options …

   🔒 Start your 14-day free trial to see the full solution →
10. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
    Q.If $z_1 = x + iy$, $z_2 = a + ib$ and $x^2 + y^2 = a^2 + b^2$, then $z_2 =$ (A) $|z_1|\, cis\left(Tan^{-1}\left(\dfrac{b}{a}\right)\right)$ (B) $|\overline{z_2}|\, z_1$ (C) $z_1\, cis\left(Tan^{-1}\left(\dfrac{y}{x}\right)\right)$ (D) $z_1$

    ›Reveal solutionSolution

    Equal moduli ($|z_1|=|z_2|$) is all that's given, so the only universally valid expression for $z_2$ is its own polar form with $|z_2|$ swapped for the equal $|z_1|$. Answer: (A).

    Concept and Intuition

    The condition $x^2+y^2=a^2+b^2$ literally states $|z_1|^2=|z_2|^2$, i.e. $|z_1|=|z_2|$ (moduli are non-negative). It says nothing about how the arguments of $z_1$ and $z_2$ relate — they can point in completely different directions. So any correct expression for $z_2$ must reduce to an identity that holds no matter what those independent arguments are; anything that mixes $z_1$'s argument into $z_2$'s value (unless it cancels out correctly) will fail for a generic choice of $a,b,x,y$.

    Step-by-Step Solution

    1. From $x^2+y^2=a^2+b^2$, conclude $|z_1|=|z_2|$.
    2. Write $z_2$ in polar form: $z_2 = |z_2|(\cos\theta + i\sin\theta) = |z_2|\,cis(\theta)$, where $\theta=\arg z_2 = Tan^{-1}(b/a)$ (for $a>0$).
    3. Since $|z_2|=|z_1|$, substitute to get $z_2 = |z_1|\,cis\!\left(Tan^{-1}(b/a)\right)$ — this is option (A), and it is just restating $z_2$ using the known-equal modulus; it holds identically.
    4. Check option (B), $|\overline{z_2}|z_1$: this equals $|z_2|\,z_1 = |z_1|z_1$, which is not generally equal to $z_2$ (e.g. $z_1=1+i$, $z_2=1-i$: $|z_1|z_1 = \sqrt2(1+i) \ne 1-i$). …

    🔒 Start your 14-day free trial to see the full solution →
11. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
    Q.If $z = (1 + 3i)^{4/3}$, then product of all the values of $z$ is (A) $28 - 96i$ (B) $28 + 96i$ (C) $\dfrac{7 + 24i}{100}$ (D) $\dfrac{7 - 24i}{100}$

    ›Reveal solutionSolution

    The problem asks for the product of all distinct values of $z = (1+3i)^{4/3}$. By writing $1+3i$ in polar form, raising to the $4/3$ power yields three distinct cube roots of $(1+3i)^4$. The product of all cube roots of a complex number equals that number itself (since the product of the three cube roots of $w$ is $w$). Hence the product is $(1+3i)^4 = -28 - 96i$, which matches option (A) after sign check — actually careful: $(1+3i)^4 = -28 - 96i$, but the options have $28-96i$ and $28+96i$; we must verify. Let’s compute: $(1+3i)^2 = 1 + 6i -9 = -8+6i$; squaring again: $(-8+6i)^2 = 64 -96i -36 = 28 -96i$. So product = $28 - 96i$, option (A).

    Tip

    The key insight: For any nonzero complex number $w$, the three cube roots are $w^{1/3}, \omega w^{1/3}, \omega^2 w^{1/3}$, where $\omega = e^{2\pi i/3}$. Their product is $w^{1/3} \cdot \omega w^{1/3} \cdot \omega^2 w^{1/3} = \omega^3 w = w$. So the product of all values of $w^{1/3}$ is $w$ itself. Here $w = (1+3i)^4$.

    Concept and Intuition

    When we have an expression like $(1+3i)^{4/3}$, it means we first raise $1+3i$ to the 4th power, then take all cube roots of that result. So $z$ is not a single number but a set of three complex numbers (the three distinct cube roots of $(1+3i)^4$). The question asks for the product of all these three values. Instead of finding each root individually and multiplying, we use a beautiful property: the product of all $n$th roots of a complex number $w$ is $(-1)^{n-1} w$. For $n=3$, this product equals $w$ (since $(-1)^{2}=1$). So the product we need is simply $(1+3i)^4$. Compute that, and we are done.

    Step-by-step solution

    1. Rewrite the expression in root form

       $z = (1+3i)^{4/3} = \big((1+3i)^4\big)^{1/3}$.

       So $z$ represents all cube roots of the complex number $w = (1+3i)^4$.

    2. Recall the property of cube roots

       For any nonzero complex number $w$, its three cube roots are $r_1, r_2, r_3$ satisfying $r_k^3 = w$. Their product is

    $r_1 r_2 r_3 = w.$

    Why? If $r_1 = \sqrt[3]{w}$ (principal), then $r_2 = \omega r_1$, $r_3 = \omega^2 r_1$, where $\omega = e^{2\pi i/3}$ satisfies $\omega^3=1$ and $1+\omega+\omega^2=0$. Then …

    🔒 Start your 14-day free trial to see the full solution →
12. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
    Q.If $z_1$ and $z_2$ are two of the $n^{th}$ roots of unity such that the line segment joining them subtends a right angle at the origin then for a positive integer $k$, $n$ takes the form (A) $4k$ (B) $4k+1$ (C) $4k+2$ (D) $4k+3$

    ›Reveal solutionSolution

    The angle at the origin between two $n^{th}$ roots of unity is always an integer multiple of $\dfrac{2\pi}{n}$; for that to equal $\dfrac{\pi}{2}$ exactly, $n$ must be a multiple of $4$.

    Concept and Intuition

    The $n^{th}$ roots of unity are the $n$ points $e^{2\pi i j/n}$, $j=0,1,\ldots,n-1$, spaced evenly at angle $\dfrac{2\pi}{n}$ apart around the unit circle. The angle subtended at the origin by the segment joining any two of them is simply the angular separation between their positions on the circle — always some integer multiple of the basic spacing $\dfrac{2\pi}{n}$.

    Step-by-Step Solution

    1. Let $z_1=e^{2\pi i j_1/n}$ and $z_2=e^{2\pi i j_2/n}$ be two $n^{th}$ roots of unity. The angle between the vectors $Oz_1$ and $Oz_2$ is $\theta=\dfrac{2\pi(j_1-j_2)}{n}$ (up to sign/direction).
    2. We require this angle to equal a right angle: $\theta=\dfrac{\pi}{2}$ (mod $2\pi$, taking the geometric angle between the rays). Writing $m=j_1-j_2$ (an integer), we need $\dfrac{2\pi m}{n}=\dfrac{\pi}{2}$ for some integer $m$, i.e. $n=4m$.
    3. Since $m$ can be chosen as any integer $1\le m< n$ making the geometry consistent, the condition for such $z_1,z_2$ to exist is precisely that $n$ is divisible by $4$. Writing $n=4k$ for a positive integer $k$ matches option (A). …

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