Q.For any two complex numbers $z_1, z_2$ and any real numbers $a, b$, $\;|az_1-bz_2|^2+|bz_1+az_2|^2=\;$ _____.

Showing the 12 most recent of 17 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.The conjugate of the multiplicative inverse of the complex number $\boldsymbol{z}=\frac{\mathbf{1}+\mathbf{7} \boldsymbol{i}}{\mathbf{3}+\boldsymbol{i}}$ is: (A) $\frac{2}{5}+\frac{1}{5} i$ (B) $\frac{1}{5}-\frac{2}{5} i$ (C) $\frac{1}{5}+\frac{2}{5} i$ (D) $1+2 i$

   ›Reveal solutionSolution

   The problem asks for the conjugate of the multiplicative inverse of a given complex number. After simplifying the number, finding its reciprocal, and then taking the conjugate, the result is $\frac{1}{5} + \frac{2}{5}i$, which corresponds to option (C).

   Concept & Intuition

   The multiplicative inverse of a complex number $z$ is $1/z$. The conjugate of a complex number $a+bi$ is $a-bi$. So we need to:

   1. Simplify $z$ to standard form $a+bi$.
   2. Compute $1/z$.
   3. Take the conjugate of that result.

   A key trick: instead of doing two separate rationalizations, we can combine steps. The conjugate of $1/z$ is actually $1/\bar{z}$ (since $\overline{1/z} = 1/\bar{z}$). This often saves work.

   ---

   Step-by-step solution

   1. Simplify $z$

   $z = \frac{1+7i}{3+i}$

   Multiply numerator and denominator by the conjugate of the denominator $(3-i)$:

   $z = \frac{(1+7i)(3-i)}{(3+i)(3-i)} = \frac{3 - i + 21i - 7i^2}{9 - i^2}$

   Since $i^2 = -1$, this becomes:

   $z = \frac{3 + 20i + 7}{9 + 1} = \frac{10 + 20i}{10} = 1 + 2i$

   2. Find the multiplicative inverse The inverse of $z = 1+2i$ is:

   $\frac{1}{z} = \frac{1}{1+2i}$

   Multiply numerator and denominator by the conjugate $1-2i$:

   $\frac{1}{1+2i} = \frac{1-2i}{(1+2i)(1-2i)} = \frac{1-2i}{1 - (2i)^2} = \frac{1-2i}{1+4} = \frac{1-2i}{5}$

   So:

   $\frac{1}{z} = \frac{1}{5} - \frac{2}{5}i$

   3. Take the conjugate …

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2. COMEDK 2026Set 2026-M1 markMCQ
   Q.

   [!FORMULA] $\text { The conjugate of } z=\frac{(\mathbf{4}+\boldsymbol{i})(\mathbf{1}-\boldsymbol{i})}{(\mathbf{1}+\boldsymbol{i})(\mathbf{2}-\boldsymbol{i})}$

   (A) $\frac{6}{5}+\frac{7}{5} i$ (B) $\frac{1}{2}-\frac{1}{2} i$ (C) $\frac{6}{5}-\frac{7}{5} i$ (D) $\frac{1}{2}+\frac{1}{2} i$

   ›Reveal solutionSolution

   Simplify $z = \dfrac{(4+i)(1-i)}{(1+i)(2-i)} = \dfrac{6}{5}-\dfrac{7}{5}i$, so $\bar{z} = \dfrac{6}{5}+\dfrac{7}{5}i$ — option (A).

   Simplify the numerator:

   $(4+i)(1-i) = 4 - 4i + i - i^2 = 4 - 3i + 1 = 5 - 3i$

   Simplify the denominator:

   $(1+i)(2-i) = 2 - i + 2i - i^2 = 2 + i + 1 = 3 + i$

   So:

   $z = \frac{5-3i}{3+i}$

   Rationalise by multiplying numerator and denominator by $3-i$:

   $z = \frac{(5-3i)(3-i)}{(3+i)(3-i)}$

   Denominator: $(3+i)(3-i) = 9 - i^2 = 10$.

   Numerator: $(5-3i)(3-i) = 15 - 5i - 9i + 3i^2 = 15 - 14i - 3 = 12 - 14i$. …

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3. COMEDK 2025Set 2025-A1 markMCQ
   Q.The complex number $\frac{1+7 i}{(2-i)^2}$ lies in (A) Quadrant 4 (B) Quadrant 3 (C) Quadrant 1 (D) Quadrant 2

   ›Reveal solutionSolution

   Simplifying gives $-1+i$: real part negative, imaginary part positive, so it lies in Quadrant 2.

   First square the denominator:

   $(2-i)^2=4-4i+i^2=4-4i-1=3-4i.$

   So the number is $\dfrac{1+7i}{3-4i}$. Multiply numerator and denominator by the conjugate $3+4i$: …

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4. COMEDK 2025Set 2025-E1 markMCQ
   Q.Given that $z$ is a real number and $z=\frac{\lambda+4 i}{1+\lambda i}$ where $\lambda \in R$, then the possible value of $\lambda$ is : (A) $-2$ (B) $2 i$ (C) $5$ (D) $\pm 2 i$

   ›Reveal solutionSolution

   The problem asks for the real parameter $\lambda$ such that $z$ is real. The key is to set the imaginary part of $z$ to zero. Solving gives $\lambda = \pm 2$, but only $\lambda = -2$ is among the options. The correct option is (A).

   ---

   We are told that $z$ is a real number, and

   $z = \frac{\lambda + 4i}{1 + \lambda i}, \quad \lambda \in \mathbb{R}.$

   Since $z$ is real, its imaginary part must be zero. The natural approach is to simplify the complex fraction by multiplying numerator and denominator by the conjugate of the denominator, then isolate the imaginary part and set it equal to zero.

   ---

   1. Multiply numerator and denominator by the conjugate of the denominator The denominator is $1 + \lambda i$; its conjugate is $1 - \lambda i$.

   $z = \frac{(\lambda + 4i)(1 - \lambda i)}{(1 + \lambda i)(1 - \lambda i)}.$

   2. Simplify the denominator

   $(1 + \lambda i)(1 - \lambda i) = 1^2 - (\lambda i)^2 = 1 - (\lambda^2 i^2) = 1 - (-\lambda^2) = 1 + \lambda^2.$

   This is a positive real number (since $\lambda$ is real).

   3. Expand the numerator

   $(\lambda + 4i)(1 - \lambda i) = \lambda(1) + \lambda(-\lambda i) + 4i(1) + 4i(-\lambda i).$

   Compute term by term:

   • $\lambda \cdot 1 = \lambda$
   • $\lambda \cdot (-\lambda i) = -\lambda^2 i$
   • $4i \cdot 1 = 4i$
   • $4i \cdot (-\lambda i) = -4\lambda i^2 = -4\lambda (-1) = 4\lambda$

   So the numerator becomes:

   $(\lambda + 4\lambda) + (-\lambda^2 i + 4i) = (5\lambda) + (4 - \lambda^2)i.$

   4. Write $z$ in standard form

   $z = \frac{5\lambda}{1 + \lambda^2} + \frac{4 - \lambda^2}{1 + \lambda^2}\, i.$

   5. Set the imaginary part to zero (since $z$ is real)

   $\frac{4 - \lambda^2}{1 + \lambda^2} = 0.$

   The denominator is always positive (never zero for real $\lambda$), so we only need the numerator to be zero:

   $$$$

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5. COMEDK 2025Set 2025-E1 markMCQ
   Q.If $z=\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^5+\left(\frac{\sqrt{3}}{2}-\frac{i}{2}\right)^5$, then (A) $\operatorname{Re}(z)>0, \operatorname{Im}(z)<0$ (B) $\operatorname{Im}(z)=0$ (C) $\operatorname{Re}(z)=0$ (D) $\operatorname{Re}(z)>0, \operatorname{Im}(z)>0$

   ›Reveal solutionSolution

   The expression simplifies to twice the real part of a complex number in polar form; because the complex numbers are conjugates, the sum is purely real, so $\operatorname{Im}(z)=0$, which corresponds to option (B).

   We are given

   $z = \left(\frac{\sqrt{3}}{2} + \frac{i}{2}\right)^5 + \left(\frac{\sqrt{3}}{2} - \frac{i}{2}\right)^5.$

   Notice that the two terms are complex conjugates: if $w = \frac{\sqrt{3}}{2} + \frac{i}{2}$, then the second term is $\overline{w}^5 = \overline{w^5}$.

   For any complex number, $w^n + \overline{w^n} = 2\,\operatorname{Re}(w^n)$, which is always a real number. Therefore $z$ is real, meaning its imaginary part is zero. That already points to option (B).

   Let’s verify by explicit computation.

   1. Recognize the polar form The modulus of $\frac{\sqrt{3}}{2} + \frac{i}{2}$ is

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

   Its argument $\theta$ satisfies

   $\cos\theta = \frac{\sqrt{3}}{2},\quad \sin\theta = \frac{1}{2} \quad\Rightarrow\quad \theta = \frac{\pi}{6}.$

   So

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

   2. Apply De Moivre’s Theorem

   $\left(e^{i\pi/6}\right)^5 = e^{i5\pi/6}.$

   Its conjugate is

   $\left(\frac{\sqrt{3}}{2} - \frac{i}{2}\right)^5 = e^{-i5\pi/6}.$

   3. Sum the two terms

   $z = e^{i5\pi/6} + e^{-i5\pi/6} = 2\cos\left(\frac{5\pi}{6}\right).$

   Since $\cos(5\pi/6) = -\frac{\sqrt{3}}{2}$, we get

   $$$$

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6. COMEDK 2025Set 2025-M1 markMCQ
   Q.If $\frac{x-1}{3+i}+\frac{y-1}{3-i}=i$ then $(y, x)=$ (A) $(-6,-4)$ (B) $(-4,-6)$ (C) $(6,-4)$ (D) $(-4,6)$

   ›Reveal solutionSolution

   The key idea is to combine the two complex fractions over a common denominator, equate real and imaginary parts, and solve the resulting linear system to find $x$ and $y$. The ordered pair $(y, x)$ is $(-4, 6)$.

   We are given

   $\frac{x-1}{3+i} + \frac{y-1}{3-i} = i$

   and need to find $(y, x)$.

   ---

   Concept and Intuition

   When an equation involves complex numbers, a powerful approach is to separate real and imaginary parts. Here, the denominators are conjugates: $3+i$ and $3-i$. Their product is a real number, so combining the fractions will simplify nicely. Once we have a single complex fraction, we can multiply both sides by the denominator, expand, and then equate the real and imaginary parts. This gives two linear equations in $x$ and $y$.

   ---

   Step-by-step solution

   1. Combine the fractions The common denominator is $(3+i)(3-i) = 3^2 - i^2 = 9 + 1 = 10$.

   $\frac{x-1}{3+i} + \frac{y-1}{3-i} = \frac{(x-1)(3-i) + (y-1)(3+i)}{10}$

   2. Expand the numerator

   $(x-1)(3-i) = 3x - 3 - xi + i$

   $(y-1)(3+i) = 3y - 3 + yi - i$

   Adding them:

   $(3x + 3y - 6) + (-x i + y i) + (i - i)$

   The $i$ and $-i$ cancel. So the numerator is

   $3x + 3y - 6 + i(y - x)$

   3. Write the equation

   $\frac{3x + 3y - 6 + i(y - x)}{10} = i$

   Multiply both sides by 10:

   $3x + 3y - 6 + i(y - x) = 10i$

   4. Equate real and imaginary parts

      Real part: $3x + 3y - 6 = 0$

      Imaginary part: $y - x = 10$

   5. Solve the system

      From the real part: $3(x + y) = 6 \Rightarrow x + y = 2$. …

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7. COMEDK 2024Set 2024-A1 markMCQ
   Q.

   [!FORMULA] $\text { The value of } \frac{i^{1004}+i^{1006}+i^{1008}+i^{1010}+i^{1012}}{i^{510}+i^{508}+i^{506}+i^{504}+i^{502}} \text { is }$

   (A) 1 (B) i (C) $-1$ (D) $-$i

   ›Reveal solutionSolution

   Reduce each power of $i$ modulo $4$: the numerator sums to $1$ and the denominator to $-1$, giving $-1$. Option (C).

   Powers of $i$ repeat with period $4$: $i^{4k}=1$, $i^{4k+2}=-1$. Reduce each exponent modulo $4$.

   Numerator:

   $i^{1004}=1,\quad i^{1006}=-1,\quad i^{1008}=1,\quad i^{1010}=-1,\quad i^{1012}=1$

   $\text{sum}=1-1+1-1+1=1$

   Denominator: …

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8. COMEDK 2024Set 2024-E1 markMCQ
   Q.

   [!FORMULA] $\text { If }(1-4 i)^3=a+i b \text { then the value of } \mathrm{a} \text { and } \mathrm{b} \text { is }$

   (A) $-47,52$ (B) $49,-74$ (C) $-74,49$ (D) $-48,-52$

   ›Reveal solutionSolution

   We expand $(1-4i)^3$ using the binomial theorem and simplify using $i^2=-1$, then separate real and imaginary parts to find $a=-47$ and $b=52$, which matches option (A).

   Concept & Intuition

   The problem asks us to cube a complex number and write the result in the standard form $a+ib$. The direct approach is to treat $1-4i$ as a binomial and expand $(1-4i)^3$ using the binomial theorem, just like $(x+y)^3$, but remembering that $i^2 = -1$ and $i^3 = -i$. This avoids any guesswork or memorization of formulas for powers of complex numbers.

   Step-by-step solution

   1. Write the cube as a binomial expansion

   $(1-4i)^3 = 1^3 + 3(1)^2(-4i) + 3(1)(-4i)^2 + (-4i)^3$

   This is the standard expansion: $(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$, with $x=1$ and $y=-4i$.

   2. Simplify each term

      • First term: $1^3 = 1$
      • Second term: $3 \cdot 1 \cdot (-4i) = -12i$
      • Third term: $3 \cdot 1 \cdot (-4i)^2 = 3 \cdot (16i^2) = 3 \cdot 16 \cdot (-1) = -48$
      • Fourth term: $(-4i)^3 = (-4)^3 \cdot i^3 = -64 \cdot (-i) = 64i$ (Recall $i^2 = -1$, $i^3 = i^2 \cdot i = -i$)

   3. Combine real and imaginary parts

      Real parts: $1 + (-48) = -47$

      Imaginary parts: $(-12i) + (64i) = 52i$

      So $(1-4i)^3 = -47 + 52i$, meaning $a = -47$ and $b = 52$. …

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9. COMEDK 2024Set 2024-M1 markMCQ
   Q.

   [!FORMULA] $\text { The modulus of the following complex number } \frac{1+i}{1-i}-\frac{1-i}{1+i} \text { is }$

   (A) 0 (B) 1 (C) 2 (D) 16

   ›Reveal solutionSolution

   The expression simplifies to a purely imaginary number $2i$, whose modulus is $2$. The correct option is (C).

   The key here is to avoid rushing into modulus calculations on the raw fractions. Instead, simplify the complex number first — that’s almost always the cleaner path. The modulus of a complex number $a+bi$ is $\sqrt{a^2+b^2}$, so once we find the simplest form, the modulus is immediate.

   1. Write the expression clearly Let

   $z = \frac{1+i}{1-i} - \frac{1-i}{1+i}.$

   Both fractions are reciprocals in disguise — notice that $\frac{1+i}{1-i}$ and $\frac{1-i}{1+i}$ are actually conjugates of each other.

   2. Simplify the first fraction Multiply numerator and denominator by the conjugate of the denominator:

   $\frac{1+i}{1-i} \cdot \frac{1+i}{1+i} = \frac{(1+i)^2}{1^2 - i^2} = \frac{1 + 2i + i^2}{1 - (-1)}.$

   Since $i^2 = -1$, we get

   $\frac{1 + 2i - 1}{2} = \frac{2i}{2} = i.$

   3. Simplify the second fraction Similarly,

   $\frac{1-i}{1+i} \cdot \frac{1-i}{1-i} = \frac{(1-i)^2}{1 - i^2} = \frac{1 - 2i + i^2}{2} = \frac{1 - 2i - 1}{2} = \frac{-2i}{2} = -i.$

   4. Combine the results $z = i - (-i) = i + i = 2i.$ …

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10. COMEDK 2023Set 2023-E1 markMCQ
    Q.If the conjugate of $(x+i y)(1-2 i)$ be $1+i$, then (A) $x=\frac{1}{5}$ (B) $y=\frac{3}{5}$ (C) $x-i y=\frac{1-i}{1+2 i}$ (D) $x+i y=\frac{1-i}{1-2 i}$

    ›Reveal solutionSolution

    Taking conjugates, $(x+iy)(1-2i)=\overline{1+i}=1-i$, hence $x+iy=\dfrac{1-i}{1-2i}=\dfrac{3+i}{5}$ (so $x=\tfrac35,\ y=\tfrac15$). This is exactly option (D).

    Given $\overline{(x+iy)(1-2i)}=1+i$. Conjugating both sides:

    $(x+iy)(1-2i)=\overline{1+i}=1-i.$

    Therefore

    $x+iy=\frac{1-i}{1-2i}=\frac{(1-i)(1+2i)}{(1-2i)(1+2i)}=\frac{1+2i-i+2}{5}=\frac{3+i}{5},$ …

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11. COMEDK 2023Set 2023-M1 markMCQ
    Q.If $z=\sqrt{3}+i$, then the argument of $z^2 e^{z-i}$ is equal to (A) $e^{\pi / 3}$ (B) $\frac{\pi}{3}$ (C) $\frac{\pi}{6}$ (D) $e^{\pi / 6}$

    ›Reveal solutionSolution

    $z-i=\sqrt3$ is a positive real, so $e^{z-i}$ contributes zero argument; $\arg(z^2)=2\arg z=\pi/3$.

    Here $z=\sqrt3+i$, so $\arg z=\tan^{-1}\!\dfrac{1}{\sqrt3}=\dfrac{\pi}{6}$ and $|z|=2$.

    Note $z-i=\sqrt3+i-i=\sqrt3$, a positive real number, so $e^{z-i}=e^{\sqrt3}>0$ has argument $0$.

    Therefore …

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12. COMEDK 2023Set 2023-M1 markMCQ
    Q.If $i=\sqrt{-1}$ and $n$ is a positive integer, then $i^n+i^{n+1}+i^{n+2}+i^{n+3}$ is equal to (A) 1 (B) $i$ (C) $i^n$ (D) 0

    ›Reveal solutionSolution

    Factor out $i^n$; the bracket $1+i+i^2+i^3=0$, so the sum is $0$.

    Factor:

    $i^n+i^{n+1}+i^{n+2}+i^{n+3}=i^n\left(1+i+i^2+i^3\right).$

    Since $i^2=-1$ and $i^3=-i$:

    $1+i+i^2+i^3=1+i-1-i=0.$ …

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