Q.The value of $\sqrt{-25}\times\sqrt{-9}$ is _____.

Showing the 12 most recent of 83 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.A point P represents a complex number $z$ in the Argand diagram. If $\omega = \dfrac{z}{\left(z - \frac{1}{3}i\right)}$ and $|\omega| = 1$, then the point P lies on (A) An ellipse (B) A circle (C) A Parabola (D) A straight line

   ›Reveal solutionSolution

   $|\omega|=1$ reduces to $|z|=|z-\frac{i}{3}|$: the locus of points equidistant from two fixed points, which is always a straight line (the perpendicular bisector).

   Concept and Intuition

   Whenever a locus condition reduces to "distance from point $A$ = distance from point $B$" (for two distinct fixed points $A,B$), the locus is the perpendicular bisector of $AB$ — a straight line. This is a very common disguise in complex-number geometry problems: an expression like $\left|\frac{z-a}{z-b}\right|=1$ always means $|z-a|=|z-b|$.

   Step-by-Step Solution

   1. $\omega=\dfrac{z}{z-\frac{i}{3}}$, and $|\omega|=1$ means $\left|\dfrac{z}{z-\frac{i}{3}}\right|=1$.
   2. Since modulus of a quotient is the quotient of moduli: $\dfrac{|z|}{|z-\frac{i}{3}|}=1 \implies |z| = \left|z-\frac{i}{3}\right|$.
   3. Writing $z=x+iy$: $|z|$ is the distance of $P=(x,y)$ from the origin $O=(0,0)$; $|z-\frac{i}{3}|$ is the distance of $P$ from the fixed point $\left(0,\frac13\right)$. …

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2. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.A, B represent the complex numbers $z_1, z_2$ in the complex plane. If $O$ is the origin not lying on $AB$, $\angle AOB = \alpha$, $OA = OB$ and $z_1, z_2$ are the roots of the equation $z^2 + pz + q = 0$, where $p, q \in \mathbb{C}$, then $p^2 =$ (A) $4q\cos^2\left(\dfrac{\alpha}{2}\right)$ (B) $4q^2\cos\left(\dfrac{\alpha}{2}\right)$ (C) $4q\sin^2\left(\dfrac{\alpha}{2}\right)$ (D) $4q^2\sin\left(\dfrac{\alpha}{2}\right)$

   ›Reveal solutionSolution

   Represent the two equal-length complex numbers with an angle $\alpha$ between them, use Vieta's formulas for the quadratic, and simplify $p^2/q$ using the half-angle identity to get $p^2=4q\cos^2(\alpha/2)$.

   Concept and Intuition

   When two complex numbers have equal modulus $r$ and a known angle $\alpha$ between their position vectors, they can be written as $re^{i\theta}$ and $re^{i(\theta+\alpha)}$ for some reference angle $\theta$. Combined with Vieta's relations for a quadratic ($z_1+z_2=-p$, $z_1z_2=q$), this lets us express $p^2/q$ purely in terms of $\alpha$ (the $\theta$ dependence cancels), using the classic identity $1+e^{i\alpha}=2\cos(\alpha/2)e^{i\alpha/2}$.

   Step-by-Step Solution

   1. Let $OA=OB=r$. Since $\angle AOB=\alpha$, write $z_1=re^{i\theta}$, $z_2=re^{i(\theta+\alpha)}$.
   2. By Vieta's formulas for $z^2+pz+q=0$: $z_1+z_2=-p$ and $z_1z_2=q$.
   3. $p = -(z_1+z_2) = -r\,e^{i\theta}\left(1+e^{i\alpha}\right)$, so $p^2 = r^2 e^{2i\theta}(1+e^{i\alpha})^2$.
   4. $q = z_1z_2 = r^2 e^{i(2\theta+\alpha)} = r^2 e^{2i\theta}e^{i\alpha}$.
   5. Divide: $\dfrac{p^2}{q} = \dfrac{(1+e^{i\alpha})^2}{e^{i\alpha}}$.
   6. Use $1+e^{i\alpha}=e^{i\alpha/2}\left(e^{-i\alpha/2}+e^{i\alpha/2}\right)=2\cos(\alpha/2)\,e^{i\alpha/2}$, so $(1+e^{i\alpha})^2 = 4\cos^2(\alpha/2)\,e^{i\alpha}$. …

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3. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.If $x = \omega^2 - \omega - 3$, $\omega$ being the non-real cube root of unity, then $x^4 + 6x^3 + 10x^2 - 12x - 19 =$ (A) $-19$ (B) $5$ (C) $12$ (D) $-14$

   ›Reveal solutionSolution

   $x=\omega^2-\omega-3$ satisfies the quadratic $x^2+6x+12=0$; reducing the quartic expression modulo this relation collapses it to the constant $5$. Answer: (B).

   Concept and Intuition

   Rather than substituting the explicit complex value of $\omega$ and grinding through complex arithmetic, it's far cleaner to find the minimal polynomial satisfied by $x$ using the cube-root-of-unity identity $\omega^2+\omega+1=0$ (equivalently $\omega^2+\omega=-1$) and $\omega^3=1$. Once we have a quadratic relation for $x$, we can reduce any higher power of $x$ using polynomial division / repeated substitution, turning a quartic evaluation into simple algebra.

   Step-by-Step Solution

   1. From $x = \omega^2-\omega-3$, write $x+3 = \omega^2-\omega$.
   2. Square both sides: $(x+3)^2 = (\omega^2-\omega)^2 = \omega^4 - 2\omega^3+\omega^2$.
   3. Use $\omega^3=1 \Rightarrow \omega^4=\omega$: so $(x+3)^2 = \omega - 2(1) + \omega^2 = (\omega^2+\omega) - 2$.
   4. Use $\omega^2+\omega = -1$: $(x+3)^2 = -1-2 = -3$.
   5. Expand: $x^2+6x+9 = -3 \Rightarrow x^2+6x+12=0$, i.e. $x^2 = -6x-12$.
   6. Compute $x^3 = x\cdot x^2 = x(-6x-12) = -6x^2-12x = -6(-6x-12)-12x = 36x+72-12x = 24x+72$.
   7. Compute $x^4 = x\cdot x^3 = x(24x+72) = 24x^2+72x = 24(-6x-12)+72x = -144x-288+72x = -72x-288$.
   8. Substitute into the target expression: …

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4. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.If $z_1 = \sqrt{3} + i\sqrt{3}$ and $z_2 = \sqrt{3} + i$, then the complex number $\left(\dfrac{z_1}{z_2}\right)^{50}$ lies in the (A) First quadrant (B) Second quadrant (C) Third quadrant (D) Fourth quadrant

   ›Reveal solutionSolution

   This tests polar representation of complex numbers and how arguments add/scale under division and powers. The resultant argument reduces to $30^\circ$, so the number lies in the first quadrant.

   Concept and Intuition

   When complex numbers are written in polar form $z=r(\cos\phi+i\sin\phi)$, division subtracts arguments and raising to a power multiplies the argument by that power. So instead of expanding $(z_1/z_2)^{50}$ algebraically, we just track the angle.

   Step-by-Step Solution

   1. Write $z_1=\sqrt3+i\sqrt3=\sqrt6\left(\cos45^\circ+i\sin45^\circ\right)$ since $\tan^{-1}\left(\frac{\sqrt3}{\sqrt3}\right)=45^\circ$ and $|z_1|=\sqrt{3+3}=\sqrt6$.
   2. Write $z_2=\sqrt3+i=2\left(\cos30^\circ+i\sin30^\circ\right)$ since $|z_2|=\sqrt{3+1}=2$ and $\tan^{-1}\left(\frac{1}{\sqrt3}\right)=30^\circ$.
   3. Then $\dfrac{z_1}{z_2}=\dfrac{\sqrt6}{2}\left(\cos15^\circ+i\sin15^\circ\right)$, argument $15^\circ$. …

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5. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.If $z_1, z_2$ are two roots of the equation $z^2 + az + b = 0$ and on the Argand plane points represented by $z_1, z_2$ and origin forms an equilateral triangle, then $a^2 =$ (A) $b$ (B) $2b$ (C) $3b$ (D) $4b$

   ›Reveal solutionSolution

   For an equilateral triangle with the origin, $z_1^2+z_2^2=z_1z_2$, which forces $a^2=3b$.

   For any equilateral triangle with vertices $z_1,z_2,z_3$,

   $z_1^2+z_2^2+z_3^2 = z_1z_2+z_2z_3+z_3z_1.$

   Here the third vertex is the origin, $z_3=0$, so this reduces to

   $z_1^2+z_2^2 = z_1z_2.$

   From $z^2+az+b=0$ (Vieta):

   $z_1+z_2=-a,\qquad z_1z_2=b.$

   Then …

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6. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.Among the roots of $3\sqrt{3}\, z^3 - i = 0$, the sum of the squares of the two roots having non-zero real part is (A) $\omega^2$ (B) $\dfrac{2}{3}$ (C) $\omega$ (D) $\dfrac{1}{3}$

   ›Reveal solutionSolution

   The two roots with non-zero real part are $\pm\tfrac12+\tfrac{i}{2\sqrt3}$; the imaginary parts cancel on squaring, leaving $\tfrac13$.

   Solve $3\sqrt3\,z^{3}-i=0$:

   $z^{3}=\frac{i}{3\sqrt3}=\frac{1}{3\sqrt3}\,e^{i\pi/2},\qquad |z|=\left(\frac{1}{3\sqrt3}\right)^{1/3}=\frac{1}{\sqrt3}.$

   The three cube roots have arguments $\dfrac{\pi/2+2k\pi}{3}$ for $k=0,1,2$:

   $\theta_0=\frac{\pi}{6},\qquad \theta_1=\frac{5\pi}{6},\qquad \theta_2=-\frac{\pi}{2}.$

   So

   $z_0=\frac{1}{\sqrt3}\!\left(\cos\tfrac{\pi}{6}+i\sin\tfrac{\pi}{6}\right)=\frac{1}{2}+\frac{i}{2\sqrt3},$

   $z_1=\frac{1}{\sqrt3}\!\left(\cos\tfrac{5\pi}{6}+i\sin\tfrac{5\pi}{6}\right)=-\frac{1}{2}+\frac{i}{2\sqrt3},$ …

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7. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.If $z_1 = 1 + i$, $z_2 = -1 - i$ represent two fixed points on the Argand plane, then the solution of $|z - z_1| = |z - z_2|$ is (A) $z = 0$ only (B) $z = 2x - ix\ (x \in \mathbb{R})$ (C) $z = x - ix\ (x \in \mathbb{R})$ (D) $z = y + iy\ (y \in \mathbb{R})$

   ›Reveal solutionSolution

   The equation $|z - z_1| = |z - z_2|$ describes the perpendicular bisector of the segment joining $z_1$ and $z_2$. For $z_1 = 1+i$ and $z_2 = -1-i$, the midpoint is $0$ and the slope of the segment is $1$, so the perpendicular bisector is the line $y = -x$, which is exactly $z = x - ix$ for real $x$. The correct option is (C).

   ---

   The key insight: The set of points equidistant from two fixed points in the plane is the perpendicular bisector of the segment joining them. This is a geometric fact that holds in the complex plane just as it does in coordinate geometry. Here, $z_1$ and $z_2$ are given, so we are finding all $z$ such that the distances to $z_1$ and $z_2$ are equal.

   Let’s work through it step by step.

   1. Write the condition algebraically. The equation $|z - z_1| = |z - z_2|$ means the distance from $z$ to $z_1$ equals the distance from $z$ to $z_2$. Let $z = x + iy$, $z_1 = 1 + i$, $z_2 = -1 - i$. Then:

   $|(x+iy) - (1+i)| = |(x+iy) - (-1-i)|$

   $|(x-1) + i(y-1)| = |(x+1) + i(y+1)|$

   2. Square both sides to remove the modulus. Squaring preserves equality because both sides are non-negative:

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

   3. Expand and simplify.

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

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

   $-2x - 2y + 2 = 2x + 2y + 2$

   Subtract 2 from both sides:

   $-2x - 2y = 2x + 2y$

   Bring terms together:

   $-4x - 4y = 0 \quad \Rightarrow \quad x + y = 0$

   4. Interpret the result. The equation $x + y = 0$ means $y = -x$. In complex form, $z = x + iy = x - ix = x(1 - i)$, where $x$ is any real number. So the solution set is all points on the line through the origin with slope $-1$. …

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8. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.Given that $\displaystyle\sum_{k=1}^{n} k(k-1) = \dfrac{n(n-1)(n+1)}{3}$ and $\omega$ and $\omega^2$ are complex cube roots of unity. If $\displaystyle\sum_{k=1}^{2026} \left(k + \dfrac{1}{\omega}\right)\left(k + \dfrac{1}{\omega^2}\right) = \dfrac{2026}{3}(N+3)$, then $N =$ (A) $2025 \times 2026$ (B) $2025 \times 2024$ (C) $2027 \times 2025$ (D) $2026 \times 2027$

   ›Reveal solutionSolution

   The sum simplifies using the identity $\frac{1}{\omega} + \frac{1}{\omega^2} = -1$ and $\frac{1}{\omega \cdot \omega^2} = 1$, turning the sum into $\sum_{k=1}^{2026} (k^2 - k + 1)$. Using the given formula, the result is $\frac{2026}{3}(2025\cdot2026 + 3)$, so $N = 2025 \times 2026$.

   Concept and intuition:

   The expression inside the sum looks like a product of two linear terms in $k$. If we expand it, we get terms involving $k^2$, $k$, and a constant. The key is that $\omega$ and $\omega^2$ are the non-real cube roots of unity, satisfying $1 + \omega + \omega^2 = 0$ and $\omega^3 = 1$. This gives simple values for sums and products of their reciprocals:

   $\frac{1}{\omega} + \frac{1}{\omega^2} = \omega^2 + \omega = -1, \quad \frac{1}{\omega} \cdot \frac{1}{\omega^2} = \frac{1}{\omega^3} = 1.$

   So the expansion becomes a clean quadratic in $k$ with no messy complex numbers.

   Step-by-step solution:

   1. Expand the product

   $\left(k + \frac{1}{\omega}\right)\left(k + \frac{1}{\omega^2}\right) = k^2 + k\left(\frac{1}{\omega} + \frac{1}{\omega^2}\right) + \frac{1}{\omega \cdot \omega^2}.$

   2. Use the properties of cube roots of unity Since $\omega^2 + \omega = -1$, we have $\frac{1}{\omega} + \frac{1}{\omega^2} = \omega^2 + \omega = -1$. Also $\frac{1}{\omega \cdot \omega^2} = \frac{1}{\omega^3} = 1$. So the expression simplifies to:

   $k^2 + k(-1) + 1 = k^2 - k + 1.$

   3. Sum over $k = 1$ to $2026$

   $\sum_{k=1}^{2026} (k^2 - k + 1) = \sum_{k=1}^{2026} k^2 - \sum_{k=1}^{2026} k + \sum_{k=1}^{2026} 1.$

   4. Apply standard summation formulas

   $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}, \quad \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}, \quad \sum_{k=1}^{n} 1 = n.$

   For $n = 2026$:

   $\sum k^2 = \frac{2026 \cdot 2027 \cdot 4053}{6}, \quad \sum k = \frac{2026 \cdot 2027}{2}, \quad \sum 1 = 2026.$

   5. Combine the sums

   $\sum (k^2 - k + 1) = \frac{2026 \cdot 2027 \cdot 4053}{6} - \frac{2026 \cdot 2027}{2} + 2026.$

   Factor $2026$:

   $= 2026 \left( \frac{2027 \cdot 4053}{6} - \frac{2027}{2} + 1 \right).$

   6. Simplify inside the parentheses Write everything with denominator 6:

   $\frac{2027 \cdot 4053}{6} - \frac{3 \cdot 2027}{6} + \frac{6}{6} = \frac{2027(4053 - 3) + 6}{6} = \frac{2027 \cdot 4050 + 6}{6}.$

   Note $4050 = 2 \cdot 2025$, so:

   $2027 \cdot 4050 = 2027 \cdot 2 \cdot 2025 = 2 \cdot 2025 \cdot 2027.$

   Thus:

   $\frac{2 \cdot 2025 \cdot 2027 + 6}{6} = \frac{2(2025 \cdot 2027 + 3)}{6} = \frac{2025 \cdot 2027 + 3}{3}.$

   7. Write the total sum

   $\sum_{k=1}^{2026} (k^2 - k + 1) = 2026 \cdot \frac{2025 \cdot 2027 + 3}{3} = \frac{2026}{3} (2025 \cdot 2027 + 3).$

   8. Compare with the given form The problem states the sum equals $\frac{2026}{3}(N + 3)$. Therefore: $N + 3 = 2025 \cdot 2027 + 3 \quad \Rightarrow \quad N = 2025 \cdot 2027.$ …

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9. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.If $x = 2 - \sqrt{3}i$, then $x^4 - 8x^3 + 16x^2 + 1 =$ (A) $14$ (B) $3x-26$ (C) $5x+32$ (D) $50$

   ›Reveal solutionSolution

   From $x=2-\sqrt{3}\,i$ we get $x^2=4x-7$; reducing the polynomial with this gives $50$ — option (D).

   Since $x-2=-\sqrt{3}\,i$, squaring gives $(x-2)^2=-3$, so

   $x^2-4x+7=0\quad\Longrightarrow\quad x^2=4x-7.$

   Reduce successive powers:

   $x^3=x\cdot x^2=x(4x-7)=4x^2-7x=4(4x-7)-7x=9x-28,$

   $x^4=x\cdot x^3=x(9x-28)=9x^2-28x=9(4x-7)-28x=8x-63.$

   Substitute into the expression: …

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10. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
    Q.If $\alpha, \beta$ are the roots of the equation $x^2 - 2\sqrt{3}x + 4 = 0$ and $0 < Arg(\alpha) < \dfrac{\pi}{2}$, then $\alpha^{2026} - \beta^{2026} =$ (A) $2^{2027}(\sqrt{3}i)$ (B) $2^{2027}i$ (C) $-2^{2026}i$ (D) $-2^{2026}(\sqrt{3}i)$

    ›Reveal solutionSolution

    The roots are $\sqrt{3}\pm i = 2e^{\pm i\pi/6}$; raising to the 2026th power gives $\alpha^{2026}-\beta^{2026} = -2^{2026}\sqrt{3}\,i$.

    Find the roots.

    $x = \frac{2\sqrt{3}\pm\sqrt{12-16}}{2} = \sqrt{3}\pm i$

    Since $0 < \operatorname{Arg}(\alpha) < \tfrac{\pi}{2}$, take $\alpha = \sqrt{3}+i$ and $\beta = \sqrt{3}-i$.

    Polar form. Both have modulus $|\alpha| = |\beta| = \sqrt{3+1} = 2$, with

    $\alpha = 2e^{i\pi/6}, \qquad \beta = 2e^{-i\pi/6}.$

    Raise to the 2026th power.

    $\alpha^{2026} = 2^{2026}e^{i\,2026\pi/6}, \qquad \frac{2026\pi}{6} = \frac{1013\pi}{3} \equiv \frac{5\pi}{3}\ (\text{mod }2\pi).$

    So $\alpha^{2026} = 2^{2026}e^{i5\pi/3}$ and $\beta^{2026} = 2^{2026}e^{-i5\pi/3}$.

    Subtract. …

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11. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
    Q.If the complex conjugate of $\dfrac{(1-i)^3}{(\sqrt3-i)^2}$ is $x+iy$, then $y - x =$ (A) 1 (B) $\dfrac{1}{2}$ (C) $2\sqrt3$ (D) $\dfrac{2}{\sqrt3}$

    ›Reveal solutionSolution

    The problem asks for $y - x$ where $x+iy$ is the complex conjugate of $\frac{(1-i)^3}{(\sqrt3-i)^2}$.

    By simplifying the numerator and denominator using polar form, then taking the conjugate, we find $x = \frac{1}{2}$ and $y = \frac{\sqrt3}{2}$, so $y - x = \frac{\sqrt3 - 1}{2}$.

    This matches option (B).

    ---

    Concept and Intuition

    When a problem involves powers of complex numbers and then a conjugate, the polar (exponential) form is your best friend.

    Why? Because multiplying and dividing in polar form is just adding and subtracting angles, and raising to a power is just multiplying the angle.

    The conjugate in polar form is simply flipping the sign of the angle.

    So instead of doing messy algebra with binomial expansions and rationalizing denominators, we can convert each complex number to its modulus and argument, do the arithmetic on angles, and then read off the real and imaginary parts at the end.

    ---

    Step-by-step solution

    1. Convert $1-i$ to polar form

    The modulus:

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

    The argument: the point $(1, -1)$ lies in the fourth quadrant, so

    $\arg(1-i) = -\frac{\pi}{4}$

    Thus

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

    2. Cube it

    $(1-i)^3 = (\sqrt{2})^3 \, e^{-i3\pi/4} = 2\sqrt{2} \, e^{-i3\pi/4}$

    3. Convert $\sqrt{3} - i$ to polar form

    Modulus:

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

    Argument: the point $(\sqrt{3}, -1)$ is in the fourth quadrant, and

    $\tan\theta = \frac{-1}{\sqrt{3}} \quad\Rightarrow\quad \theta = -\frac{\pi}{6}$

    So

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

    4. Square the denominator

    $(\sqrt{3} - i)^2 = (2)^2 \, e^{-i\pi/3} = 4 \, e^{-i\pi/3}$

    5. Form the fraction

    $\frac{(1-i)^3}{(\sqrt{3} - i)^2} = \frac{2\sqrt{2} \, e^{-i3\pi/4}}{4 \, e^{-i\pi/3}} = \frac{\sqrt{2}}{2} \, e^{-i(3\pi/4 - \pi/3)}$

    Simplify the exponent:

    $\frac{3\pi}{4} - \frac{\pi}{3} = \frac{9\pi - 4\pi}{12} = \frac{5\pi}{12}$

    So

    $\frac{(1-i)^3}{(\sqrt{3} - i)^2} = \frac{\sqrt{2}}{2} \, e^{-i5\pi/12}$

    6. Take the complex conjugate

    The conjugate of $re^{i\theta}$ is $re^{-i\theta}$. Here we have $e^{-i5\pi/12}$, so its conjugate is $e^{i5\pi/12}$.

    Thus

    $x+iy = \frac{\sqrt{2}}{2} \, e^{i5\pi/12}$

    7. Convert to rectangular form

    We need $\cos(5\pi/12)$ and $\sin(5\pi/12)$.

    Note that $5\pi/12 = 75^\circ = 45^\circ + 30^\circ$.

    $$\cos 75^\circ = \cos(45^\circ+30^\circ) = \cos45^\circ\cos30^\circ - \sin45^\circ\sin30^\circ …$$

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12. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
    Q.If $Arg\left(\dfrac{z-1}{z+5}\right) = \pi/3$ and $|z-1| = |z+5|$, then $|z|^2 =$ (A) $36$ (B) $31$ (C) $41$ (D) $39$

    ›Reveal solutionSolution

    The condition $|z-1| = |z+5|$ forces $z$ to lie on the perpendicular bisector of the segment joining $1$ and $-5$, which is the vertical line $x = -2$. The argument condition $\operatorname{Arg}\!\left(\frac{z-1}{z+5}\right) = \pi/3$ means the angle between the vectors $z-1$ and $z+5$ is $60^\circ$. Solving these together gives $z = -2 \pm 3\sqrt{3}\,i$, so $|z|^2 = 31$.

    ---

    Concept and Intuition

    We have two geometric facts about a complex number $z$:

    1. $|z-1| = |z+5|$ means $z$ is equidistant from the points $1$ and $-5$ in the complex plane. That’s the perpendicular bisector of the segment joining them — a vertical line through their midpoint.
    2. $\operatorname{Arg}\!\left(\frac{z-1}{z+5}\right) = \pi/3$ means the directed angle from the vector $z+5$ to the vector $z-1$ is $60^\circ$. In other words, the triangle formed by $z$, $1$, and $-5$ has a fixed angle at $z$.

    Combining these gives a unique location for $z$, and then $|z|^2$ is just the squared distance from the origin.

    ---

    Step-by-step solution

    1. Interpret the modulus condition

       $|z-1| = |z+5|$ says $z$ is equidistant from $1$ and $-5$. The midpoint is $\frac{1 + (-5)}{2} = -2$, and the segment is horizontal, so the perpendicular bisector is the vertical line $x = -2$.

       Hence $z = -2 + iy$ for some real $y$.

    2. Interpret the argument condition

       For $z = -2 + iy$, compute:

    $z-1 = -3 + iy, \qquad z+5 = 3 + iy.$

    The argument of the quotient is the difference of arguments:

    $\operatorname{Arg}\!\left(\frac{z-1}{z+5}\right) = \operatorname{Arg}(z-1) - \operatorname{Arg}(z+5) = \frac{\pi}{3}.$

    So the angle between the vectors $(-3, y)$ and $(3, y)$ is $60^\circ$.

    3. Use the tangent of the angle difference Let $\theta_1 = \operatorname{Arg}(-3 + iy)$ and $\theta_2 = \operatorname{Arg}(3 + iy)$. Then

    $\tan(\theta_1 - \theta_2) = \frac{\tan\theta_1 - \tan\theta_2}{1 + \tan\theta_1 \tan\theta_2}.$

    Here $\tan\theta_1 = \frac{y}{-3}$ and $\tan\theta_2 = \frac{y}{3}$. So

    $\tan(\theta_1 - \theta_2) = \frac{\frac{y}{-3} - \frac{y}{3}}{1 + \left(\frac{y}{-3}\right)\left(\frac{y}{3}\right)} = \frac{-\frac{2y}{3}}{1 - \frac{y^2}{9}} = \frac{-2y/3}{(9 - y^2)/9} = \frac{-6y}{9 - y^2}.$

    Since $\theta_1 - \theta_2 = \pi/3$, we have $\tan(\pi/3) = \sqrt{3}$. Thus

    $\frac{-6y}{9 - y^2} = \sqrt{3}.$

    4. Solve for $y$ Multiply: $-6y = \sqrt{3}(9 - y^2)$. Rearranging:

    $\sqrt{3}\,y^2 - 6y - 9\sqrt{3} = 0.$

    Divide by $\sqrt{3}$:

    $y^2 - \frac{6}{\sqrt{3}}y - 9 = 0 \quad\Rightarrow\quad y^2 - 2\sqrt{3}\,y - 9 = 0.$

    Solve:

    $$$$

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