Q.If $(x+iy)^{\frac{1}{3}} = (a+ib)$, then prove that $\left(\dfrac{x}{a} + \dfrac{y}{b}\right) = 4(a^2 - b^2)$. OR If $(a+ib) = \dfrac{c+i}{c-i}$, where $c$ is real. Prove that $a^2 + b^2 = 1$ and $\dfrac{b}{a} = \dfrac{2c}{c^2 - 1}$.

Showing the 12 most recent of 94 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.The multiplicative inverse of $\sqrt{5} + 3i$ is

   (a) $\sqrt{5} - 3i$

   (b) $\dfrac{\sqrt{5}}{14} + i \cdot \dfrac{3}{14}$

   (c) $\dfrac{\sqrt{5}}{14} - i \cdot \dfrac{3}{14}$

   (d) $3 - \sqrt{5}i$

   ›Reveal solutionSolution

   The multiplicative inverse of $a+ib$ is $\dfrac{a-ib}{a^2+b^2}$; here $a=\sqrt5, b=3$.

   The multiplicative inverse of $z = \sqrt5 + 3i$ is $\dfrac{1}{z}$. Rationalise by multiplying by the conjugate $\sqrt5 - 3i$: …

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2. CBSE 2026Set ANNUAL1 markMCQ
   Q.The modulus of $\dfrac{1+i}{1-i} - \dfrac{1-i}{1+i}$ is

   (a) 2

   (b) -2

   (c) 1

   (d) -1

   ›Reveal solutionSolution

   Simplify each fraction using the conjugate, subtract, then take the modulus of the resulting purely imaginary number.

   $\dfrac{1+i}{1-i} = \dfrac{(1+i)^2}{(1-i)(1+i)} = \dfrac{1+2i+i^2}{1-i^2} = \dfrac{1+2i-1}{1+1} = \dfrac{2i}{2} = i$ …

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3. CBSE 2026Set ANNUAL1 markMCQ
   Q.If $\left(\dfrac{1+i}{1-i}\right)^m = 1$ then the least integral value of $m$ is

   (a) 1

   (b) 2

   (c) 3

   (d) 4

   ›Reveal solutionSolution

   Simplify the base to $i$, then find the smallest positive integer power of $i$ equal to 1 (the order of $i$ is 4).

   As shown above, $\dfrac{1+i}{1-i} = i$. So the equation becomes $i^m = 1$.

   …

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4. CBSE 2026Set ANNUAL1 markMCQ
   Q.The value of $i^{28}$ is —

   (a) $i$

   (b) $-i$

   (c) $1$

   (d) $-1$

   ›Reveal solutionSolution

   $i^{28} = 1$, option (c).

   Recall $i^2=-1$, so $i^4 = (i^2)^2 = (-1)^2 = 1$. Powers of $i$ cycle with period 4: $i, -1, -i, 1, i, -1, \dots$

   …

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5. CBSE 2026Set ANNUAL1 mark
   Q.The conjugate of the complex number $2 - 3i$ is ______.

   ›Reveal solutionSolution

   The conjugate of $2-3i$ is $2+3i$.

   For a complex number $z=a+ib$, the conjugate $\bar z = a - ib$ — the sign of the imaginary part is reversed while the real part stays unchanged.

   …

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6. CBSE 2026Set ANNUAL1 markMCQ
   Q.Multiplicative inverse of 1 + √3i is:

   (a) 1 - √3i

   (b) (1 - √3i)/2

   (c) (1 + √3i)/4

   (d) None of these

   ›Reveal solutionSolution

   z⁻¹ = conjugate(z)/|z|²; for z = 1+√3i this gives (1−√3i)/4, which isn't among options (a)-(c) as written, so the correct choice is (d).

   For a complex number $z = 1 + \sqrt{3}i$, the multiplicative inverse is:

   $z^{-1} = \frac{\bar z}{|z|^2}$

   Here $\bar z = 1 - \sqrt{3}i$, and $|z|^2 = 1^2 + (\sqrt{3})^2 = 1 + 3 = 4$.

   So:

   $z^{-1} = \frac{1-\sqrt3 i}{4}$

   …

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7. CBSE 2026Set ANNUAL1 mark
   Q.The modulus of (1 + i)/(1 - i) is ..............

   ›Reveal solutionSolution

   Rationalize (1+i)/(1-i) to a+bi form, or use |z1/z2| = |z1|/|z2| — both give modulus 1.

   Method using the modulus quotient rule $\left|\dfrac{z_1}{z_2}\right| = \dfrac{|z_1|}{|z_2|}$:

   $|1+i| = \sqrt{1^2+1^2}=\sqrt2, \qquad |1-i|=\sqrt{1^2+(-1)^2}=\sqrt2$

   $\left|\frac{1+i}{1-i}\right| = \frac{\sqrt2}{\sqrt2} = 1$

   …

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8. CBSE 2026Set ANNUAL1 markMCQ
   Q.$i^9 \cdot i^{19}$ is equal to

   (a) $-1$

   (b) $-i$

   (c) $1$

   (d) $0$

   ›Reveal solutionSolution

   Using $i^4=1$, reduce the exponents mod 4: $i^9 = i$ and $i^{19}=i^3=-i$, so the product is $i\cdot(-i) = 1$.

   Recall $i^2=-1$, and powers of $i$ repeat with period 4: $i^1=i,\ i^2=-1,\ i^3=-i,\ i^4=1$.

   $i^9 = i^{4\times2+1} = (i^4)^2 \cdot i = 1\cdot i = i$.

   …

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9. CBSE 2026Set ANNUAL1 markMCQ
   Q.Conjugate of $3 + 5i$ is:

   (a) $-3+5i$

   (b) $-3-5i$

   (c) $3-5i$

   (d) $3+i$

   ›Reveal solutionSolution

   The conjugate of $z=a+ib$ is $\bar{z}=a-ib$; only the sign of the imaginary part changes.

   Here $z = 3+5i$, so $a=3$, $b=5$.

   …

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10. CBSE 2026Set ANNUAL1 mark
    Q.Evaluate $i^9 + i^{19}$.

    ›Reveal solutionSolution

    Using $i^4=1$, reduce each exponent mod 4: $i^9=i$ and $i^{19}=-i$, which sum to 0.

    Since $i^4=1$, any power $i^k = i^{(k \bmod 4)}$.

    $i^9 = i^{(4\times2+1)} = (i^4)^2\cdot i^1 = 1\cdot i = i$.

    …

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11. CBSE 2026Set ANNUAL1 mark
    Q.Write modulus of $5 - 3i$.

    ›Reveal solutionSolution

    Modulus of $a+ib$ is $\sqrt{a^2+b^2}$; here that gives $\sqrt{34}$.

    For $z=5-3i$, $a=5$, $b=-3$.

    …

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12. CBSE 2026Set 1A1 markMCQ
    Q.The value of $\sqrt{-25} \times \sqrt{-9}$ is -

    (1) $15$

    (2) $-15$

    (3) $15i$

    (4) None of these

    ›Reveal solutionSolution

    Write $\sqrt{-25}=5i$ and $\sqrt{-9}=3i$; their product is $15i^2=-15$.

    The rule $\sqrt{a}\,\sqrt{b}=\sqrt{ab}$ fails for negative numbers, so convert first:

    $\sqrt{-25}=5i,\qquad \sqrt{-9}=3i.$ …

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