Q.Express the following in the form of $a + bi$:

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 $\sqrt5 - 3i$:

   $\dfrac{1}{\sqrt5+3i} = \dfrac{\sqrt5-3i}{(\sqrt5+3i)(\sqrt5-3i)} = \dfrac{\sqrt5-3i}{(\sqrt5)^2 - (3i)^2} = \dfrac{\sqrt5-3i}{5-(-9)} = \dfrac{\sqrt5-3i}{14}$

   $= \dfrac{\sqrt5}{14} - i\cdot\dfrac{3}{14}$

   ✓Final answer

   (c) $\dfrac{\sqrt5}{14} - i\cdot\dfrac{3}{14}$.

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.

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

   Similarly, $\dfrac{1-i}{1+i}$ is the conjugate expression, giving $-i$.

   So the expression becomes:

   $i - (-i) = 2i$

   Modulus of $2i$ is $|2i| = \sqrt{0^2+2^2} = 2$.

   ✓Final answer

   (a) 2.

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 $i^m = 1$.

   The powers of $i$ cycle with period 4: $i^1=i,\ i^2=-1,\ i^3=-i,\ i^4=1$, and then repeat. The smallest positive integer $m$ for which $i^m = 1$ is $m = 4$.

   ✓Final answer

   (d) 4.

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: $i, -1, -i, 1, i, -1, \dots$

   Since $28 = 4 \times 7$ is exactly divisible by 4, $i^{28} = (i^4)^7 = 1^7 = 1$.

   ✓Final answer

   The correct option is (c) $1$.

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.

   Here $z = 2-3i$, so $a=2$, $b=-3$. Its conjugate is $\bar z = 2-(-3)i = 2+3i$.

   ✓Final answer

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

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:

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

   Check against the given options: (a) $1-\sqrt3i$ — wrong scale. (b) $\dfrac{1-\sqrt3i}{2}$ — wrong denominator (should be 4, not 2). (c) $\dfrac{1+\sqrt3i}{4}$ — correct denominator but wrong sign (this is $z$ itself divided by 4, not $\bar z$ divided by 4). None of (a), (b), (c) equal the correct value $\dfrac{1-\sqrt3i}{4}$.

   ✓Final answer

   The correct inverse is $\dfrac{1-\sqrt3 i}{4}$, which is not listed exactly among (a)-(c) — option (d) None of these.

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 $\left|\dfrac{z_1}{z_2}\right| = \dfrac{|z_1|}{|z_2|}$:

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

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

   (Check by rationalizing: $\dfrac{1+i}{1-i}\times\dfrac{1+i}{1+i} = \dfrac{(1+i)^2}{2} = \dfrac{2i}{2}=i$, and $|i|=1$ ✓.)

   ✓Final answer

   The modulus of $\dfrac{1+i}{1-i}$ is $1$.

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: $i^1=i,\ i^2=-1,\ i^3=-i,\ i^4=1$.

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

   $i^{19} = i^{4\times4+3} = (i^4)^4 \cdot i^3 = 1 \cdot (-i) = -i$.

   So $i^9 \cdot i^{19} = i \cdot (-i) = -i^2 = -(-1) = 1$.

   ✓Final answer

   $i^9 \cdot i^{19} = 1$, which is option (c).

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 $a=3$, $b=5$.

   By definition, $\bar{z} = a - ib = 3-5i$.

   ✓Final answer

   The correct option is (c) $3-5i$.

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 $i^k = i^{(k \bmod 4)}$.

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

    $i^{19} = i^{(4\times4+3)} = (i^4)^4\cdot i^3 = 1\cdot(-i) = -i$.

    Adding: $i^9+i^{19} = i + (-i) = 0$.

    ✓Final answer

    $i^9+i^{19}=0$.

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

    $|z| = \sqrt{a^2+b^2} = \sqrt{5^2+(-3)^2} = \sqrt{25+9} = \sqrt{34}$.

    ✓Final answer

    $|5-3i| = \sqrt{34}$.

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:

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

    Then $\sqrt{-25}\times\sqrt{-9}=5i\times 3i=15i^2=15(-1)=-15.$

    ✓Final answer

    The value is $-15$, i.e. option (2).