Q.Reduce $\left(\dfrac{1}{1 - 4i} - \dfrac{2}{1 + i}\right)\left(\dfrac{3 - 4i}{5 + i}\right)$ to the standard form.

Showing the 12 most recent of 59 on this concept.

1. KEAM 2026Set eng-2026-04174 marksMCQ
   Q.If the complex number $z=x+iy$ satisfies the equation $5z-2\bar{z}=\dfrac{7-7i}{1+i}$, then the value of $x+y$ is equal to (A) 7 (B) -3 (C) 3 (D) -1 (E) -7

   ›Reveal solutionSolution

   Simplify the right side to $-7i$, equate real and imaginary parts, get $x=0,\;y=-1$, so $x+y=-1$.

   Simplify the right-hand side:

   $\frac{7-7i}{1+i}=\frac{(7-7i)(1-i)}{(1+i)(1-i)}=\frac{7-7i-7i+7i^2}{2}=\frac{-14i}{2}=-7i.$

   With $z=x+iy$, $\bar z=x-iy$:

   $5z-2\bar z=5(x+iy)-2(x-iy)=3x+7iy.$

   Setting $3x+7iy=-7i$ gives $3x=0\Rightarrow x=0$ and $7y=-7\Rightarrow y=-1$.

   Thus $x+y=-1$.

   ✓Final answer

   The correct option is (D).

2. KEAM 2026Set eng-2026-04174 marksMCQ
   Q.Let $z=1+i$, where $i=\sqrt{-1}$. If $z-\dfrac{24\bar{z}}{z^2}=\lambda z$, then the value of $\lambda$ is equal to (A) 12 (B) 13 (C) 18 (D) 23 (E) 24

   ›Reveal solutionSolution

   Evaluate $\dfrac{24\bar z}{z^2}=-12-12i$; subtracting from $z$ gives $13(1+i)=13z$, so $\lambda=13$.

   For $z=1+i$: $z^2=(1+i)^2=2i$ and $\bar z=1-i$. Then

   $\frac{24\bar z}{z^2}=\frac{24(1-i)}{2i}=\frac{12(1-i)}{i}=12(1-i)(-i)=12(-i+i^2)=-12-12i.$

   Therefore

   $z-\frac{24\bar z}{z^2}=(1+i)-(-12-12i)=13+13i=13(1+i)=13z.$

   Comparing with $\lambda z$ gives $\lambda=13$.

   ✓Final answer

   The correct option is (B).

3. KEAM 2026Set eng-2026-04174 marksMCQ
   Q.The complex number $z$ satisfying the equation $\dfrac{Re(z)}{2+i}+\dfrac{Im(z)}{1+2i}=\dfrac{3}{1-2i}$, is (A) $6-5i$ (B) $-4+5i$ (C) $5-4i$ (D) $3-5i$ (E) $4-5i$

   ›Reveal solutionSolution

   Rationalize each term and match real/imaginary parts to solve $2a+b=3,\ -a-2b=6$, giving $z=4-5i$.

   Let $\mathrm{Re}(z)=a$, $\mathrm{Im}(z)=b$ (both real). Rationalizing:

   $\frac{a}{2+i}=\frac{a(2-i)}{5},\quad \frac{b}{1+2i}=\frac{b(1-2i)}{5},\quad \frac{3}{1-2i}=\frac{3(1+2i)}{5}.$

   Multiplying through by $5$:

   $a(2-i)+b(1-2i)=3(1+2i).$

   Real part: $2a+b=3$. Imaginary part: $-a-2b=6$.

   From the first, $b=3-2a$; substituting: $-a-2(3-2a)=6\Rightarrow 3a-6=6\Rightarrow a=4$, then $b=3-8=-5$.

   So $z=4-5i$.

   ✓Final answer

   The correct option is (E).

4. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.Let $z=\dfrac{a-\frac{i}{2}}{i-2}$, where $a$ is a real number and $i=\sqrt{-1}$. If $\mathrm{Im}(z)=0$, then the value of $a$ is equal to (A) 1 (B) 2 (C) 3 (D) 4 (E) 0

   ›Reveal solutionSolution

   Multiply by the conjugate; setting the imaginary part to zero gives $a=1$.

   $z=\dfrac{a-\frac{i}{2}}{i-2}$. Multiply numerator and denominator by $\overline{(i-2)}=(-2-i)$; denominator $=(-2)^2+1^2=5$.

   Numerator:

   $\left(a-\tfrac{i}{2}\right)(-2-i)=(-2a-\tfrac12)+i(1-a).$

   So $\mathrm{Im}(z)=\dfrac{1-a}{5}$. Setting it to $0$:

   $1-a=0\Rightarrow a=1.$

   ✓Final answer

   The correct option is (A).

5. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.The value of $\left[\dfrac{5i}{(3+i)(3-i)}\right]^{2026}$ is equal to (A) $\dfrac{1}{2^{2026}}$ (B) $\dfrac{1}{2^{1013}}$ (C) $\dfrac{-1}{2^{1013}}$ (D) $\dfrac{i}{2^{2026}}$ (E) $\dfrac{-1}{2^{2026}}$

   ›Reveal solutionSolution

   Base simplifies to $i/2$; $i^{2026}=-1$, giving $\dfrac{-1}{2^{2026}}$.

   $(3+i)(3-i)=9-i^2=10$, so

   $\dfrac{5i}{(3+i)(3-i)}=\dfrac{5i}{10}=\dfrac{i}{2}.$

   Then

   $\left(\dfrac{i}{2}\right)^{2026}=\dfrac{i^{2026}}{2^{2026}}.$

   Since $2026=4\cdot506+2$, $i^{2026}=i^2=-1$. Thus the value is $\dfrac{-1}{2^{2026}}$.

   ✓Final answer

   The correct option is (E).

6. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.If $z|z|=24+7i$, where $z$ is a complex number, then the value of $|z|$ is equal to (A) 5 (B) 7 (C) 12 (D) 15 (E) 25

   ›Reveal solutionSolution

   $|z|\cdot|z|=|24+7i|=25\Rightarrow|z|=5$.

   Take the modulus of both sides of $z|z|=24+7i$. Since $|z|$ is real and positive,

   $|z|\cdot|z|=|24+7i|=\sqrt{24^2+7^2}=\sqrt{576+49}=\sqrt{625}=25.$

   So $|z|^2=25\Rightarrow|z|=5$.

   ✓Final answer

   The correct option is (A).

7. KEAM 2026Set eng-2026-04194 marksMCQ
   Q.Let $z_1 = \frac{5 + 7i}{7 - 5i}, z_2 = \frac{3 + 2i}{3 - 2i}$ and $z_3 = \frac{1 + 11i}{11 - i}$. Then $z_1\overline{z_1} + z_2\overline{z_2} + z_3\overline{z_3}$ is equal to (A) 2 (B) $1+2i$ (C) 1 (D) 3 (E) $1-2i$

   ›Reveal solutionSolution

   $z\overline{z}=|z|^2$, and each fraction has numerator and denominator of equal modulus.

   For a complex number, $z\overline{z} = |z|^2$, and $\left|\frac{p}{q}\right|^2 = \frac{|p|^2}{|q|^2}$.

   • $z_1=\frac{5+7i}{7-5i}$: $\frac{25+49}{49+25}=\frac{74}{74}=1$.
   • $z_2=\frac{3+2i}{3-2i}$: $\frac{9+4}{9+4}=1$.
   • $z_3=\frac{1+11i}{11-i}$: $\frac{1+121}{121+1}=\frac{122}{122}=1$.

   Sum $= 1+1+1 = 3$.

   ✓Final answer

   The correct option is (D).

8. KEAM 2026Set eng-2026-04204 marksMCQ
   Q.Given that $i^2 = -1$. If $z_1 = (7 + i\sqrt{5})^2 + (7 - i\sqrt{5})^2$ and $z_2 = (3 + 2i)^3 - (3 - 2i)^3$, then (A) $z_1$ is a purely imaginary number and $z_2$ is a purely real number (B) $z_1$ is a purely real number and $z_2$ is a purely imaginary number (C) both $z_1$ and $z_2$ are purely imaginary numbers (D) both $z_1$ and $z_2$ are purely real numbers (E) $z_1 + z_2$ is a purely real number

   ›Reveal solutionSolution

   A number plus its conjugate is real; a number minus its conjugate is imaginary. $z_1$ is a sum of conjugates so it is purely real, while $z_2$ is a difference of conjugates so it is purely imaginary.

   Let $w=(7+i\sqrt5)^2$. Then $(7-i\sqrt5)^2=\overline{w}$, so

   $z_1 = w + \overline{w} = 2\,\operatorname{Re}(w),$

   which is a purely real number.

   (Explicitly $w=49-5+14i\sqrt5=44+14i\sqrt5$, so $z_1=88$.)

   Let $u=(3+2i)^3$. Then $(3-2i)^3=\overline{u}$, so

   $z_2 = u - \overline{u} = 2i\,\operatorname{Im}(u),$

   which is a purely imaginary number.

   (Explicitly $u=27+54i+36i^2+8i^3=-9+46i$, so $z_2=2i\cdot46=92i$.)

   Hence $z_1$ is purely real and $z_2$ is purely imaginary.

   ✓Final answer

   The correct option is (B).

9. KEAM 2026Set eng-2026-04204 marksMCQ
   Q.If $z_1 = 1 + 3i$, $z_2 = -3i + 5$, then $\left(z_1 \overline{z_2} + z_2 \overline{z_1}\right) + \overline{\left(z_1 \overline{z_2} + z_2 \overline{z_1}\right)}$ is equal to (A) $-16$ (B) $1 + i$ (C) $1$ (D) $1 - i$ (E) $-16i$

   ›Reveal solutionSolution

   The expression $z_1\overline{z_2}+z_2\overline{z_1}$ equals $2\operatorname{Re}(z_1\overline{z_2})=-8$, which is real. Adding its own conjugate (also $-8$) gives $-16$.

   With $z_1=1+3i$ and $z_2=5-3i$, so $\overline{z_2}=5+3i$.

   Compute $z_1\overline{z_2}=(1+3i)(5+3i)=5+3i+15i+9i^2=5+18i-9=-4+18i$.

   Note $z_2\overline{z_1}=\overline{z_1\overline{z_2}}=-4-18i$, so

   $z_1\overline{z_2}+z_2\overline{z_1}=(-4+18i)+(-4-18i)=-8,$

   a real number (as it must be, being $2\operatorname{Re}(z_1\overline{z_2})$).

   Call this $w=-8$. The required quantity is $w+\overline{w}=-8+(-8)=-16$.

   ✓Final answer

   The correct option is (A).

10. KEAM 2026Set eng-2026-04214 marksMCQ
    Q.If $|z + 4| = 2|z + 1|$, where $z$ is a complex number, then $|z|$ is equal to (A) 0 (B) 2 (C) 4 (D) 8 (E) 16

    ›Reveal solutionSolution

    Squaring the modulus condition gives the circle $x^2+y^2=4$, so $|z|=2$.

    Let $z=x+iy$. Then $|z+4|=2|z+1|$ squared:

    $(x+4)^2+y^2=4\big[(x+1)^2+y^2\big].$

    Expand: $x^2+8x+16+y^2=4x^2+8x+4+4y^2$, giving $0=3x^2+3y^2-12$, i.e. $x^2+y^2=4$.

    Hence $|z|=\sqrt{x^2+y^2}=2$.

    ✓Final answer

    The correct option is (B).

11. KEAM 2026Set eng-2026-04214 marksMCQ
    Q.If $z(3 - i) = 2 + i$, then $z^2 =$ (A) $\dfrac{i}{2}$ (B) $\dfrac{-i}{2}$ (C) $\dfrac{1}{2}$ (D) $\dfrac{-1}{2}$ (E) $\dfrac{1 + i}{2}$

    ›Reveal solutionSolution

    $z=\frac{1+i}{2}$ and squaring gives $\frac{i}{2}$.

    From $z(3-i)=2+i$:

    $z=\frac{2+i}{3-i}=\frac{(2+i)(3+i)}{(3-i)(3+i)}=\frac{6+2i+3i-1}{10}=\frac{5+5i}{10}=\frac{1+i}{2}.$

    Then

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

    ✓Final answer

    The correct option is (A).

12. KEAM 2026Set eng-2026-04214 marksMCQ
    Q.The imaginary part of $\dfrac{1 - i\sqrt{3}}{1 + i\sqrt{3}}$ is (A) $\dfrac{-1}{2}$ (B) $\dfrac{1}{2}$ (C) $\dfrac{\sqrt{3}}{2}$ (D) $\dfrac{-\sqrt{3}}{2}$ (E) $\dfrac{\sqrt{3}}{4}$

    ›Reveal solutionSolution

    Rationalizing gives $\frac{-1-i\sqrt3}{2}$, whose imaginary part is $-\frac{\sqrt3}{2}$.

    Multiply numerator and denominator by the conjugate $1-i\sqrt3$:

    $\frac{(1-i\sqrt3)^2}{(1+i\sqrt3)(1-i\sqrt3)}=\frac{1-2i\sqrt3+(i\sqrt3)^2}{1+3}=\frac{1-2i\sqrt3-3}{4}=\frac{-2-2i\sqrt3}{4}=\frac{-1-i\sqrt3}{2}.$

    The imaginary part is $-\dfrac{\sqrt3}{2}$.

    ✓Final answer

    The correct option is (D).