Q.Solve the following equation for $x,y\in\mathbb{R}$ : $(4-5i)x+(2+3i)y = 10-7i$

Showing the 12 most recent of 13 on this concept.

1. CBSE 2026Set 2A2 marks
   Q.Find a square root of the complex number $3+4i$.

   ›Reveal solutionSolution

   Write $\sqrt{3+4i}=x+iy$, square both sides, match real/imaginary parts, and use the modulus as a third equation to pin down $x,y$.

   Let $\sqrt{3+4i}=x+iy$ where $x,y$ are real. Squaring,

   $x^2-y^2+2ixy = 3+4i.$

   Comparing real and imaginary parts:

   $x^2-y^2=3 \qquad\text{and}\qquad 2xy=4 \Rightarrow xy=2.$

   A third relation comes from equating moduli, $|x+iy|^2=|3+4i|$:

   $x^2+y^2=\sqrt{3^2+4^2}=\sqrt{25}=5.$

   …

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2. CBSE 2026Set 2A2 marks
   Q.If the Arg $\bar{z}_1$ and Arg $z_2$ are $\frac{\pi}{5}$ and $\frac{\pi}{3}$ respectively, then find (Arg $z_1$ + Arg $z_2$).

   ›Reveal solutionSolution

   Use $\text{Arg}(\bar z)=-\text{Arg}(z)$ to get $\text{Arg}\,z_1$ from $\text{Arg}\,\bar z_1$, then add.

   For any complex number $z=r\,\text{cis}\,\theta$, its conjugate is $\bar z=r\,\text{cis}(-\theta)$, so

   $\text{Arg}(\bar z)=-\text{Arg}(z).$

   Given $\text{Arg}\,\bar z_1=\dfrac{\pi}{5}$, this means $-\text{Arg}\,z_1=\dfrac{\pi}{5}$, i.e. $\text{Arg}\,z_1=-\dfrac{\pi}{5}$.

   …

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3. CBSE 2025Set 2A2 marks
   Q.Write the complex number $\dfrac{4+3i}{(2+3i)(4-3i)}$ in the form $a+ib$.

   ›Reveal solutionSolution

   Simplify the denominator first, then rationalise by multiplying by its conjugate.

   First simplify the denominator $(2+3i)(4-3i)$:

   $(2+3i)(4-3i) = 8 - 6i + 12i - 9i^2 = 8 + 6i + 9 = 17+6i,$

   using $i^2=-1$ so $-9i^2=9$.

   So the expression becomes $\dfrac{4+3i}{17+6i}$. Multiply numerator and denominator by the conjugate $17-6i$:

   $\frac{4+3i}{17+6i}\cdot\frac{17-6i}{17-6i} = \frac{(4+3i)(17-6i)}{17^2+6^2}.$

   …

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4. CBSE 2025Set 2A2 marks
   Q.Write $z = -\sqrt{7} + i\sqrt{21}$ in the polar form.

   ›Reveal solutionSolution

   Find the modulus $r=|z|$ and argument $\theta$ from $\cos\theta=x/r,\ \sin\theta=y/r$, then write $z=r(\cos\theta+i\sin\theta)$.

   Given $z = -\sqrt7 + i\sqrt{21}$, so $x=-\sqrt7$, $y=\sqrt{21}$.

   Modulus:

   $r = \sqrt{x^2+y^2} = \sqrt{7+21} = \sqrt{28} = 2\sqrt7.$

   Argument: since $x<0$ and $y>0$, $z$ lies in the second quadrant.

   $\cos\theta = \frac{x}{r} = \frac{-\sqrt7}{2\sqrt7} = -\frac12, \qquad \sin\theta = \frac{y}{r} = \frac{\sqrt{21}}{2\sqrt7} = \frac{\sqrt3}{2}.$

   …

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5. CBSE 2023Set 2A2 marks
   Q.Find the square root of the complex number $7 + 24i$.

   ›Reveal solutionSolution

   Write $\sqrt{7+24i}=x+iy$, square both sides, match real and imaginary parts, and solve for $x,y$ using the modulus as a third equation.

   Let $\sqrt{7+24i} = x+iy$ where $x,y$ are real. Squaring,

   $x^2 - y^2 + 2ixy = 7+24i.$

   Comparing real and imaginary parts:

   $x^2-y^2 = 7 \qquad \text{and} \qquad 2xy = 24 \Rightarrow xy = 12.$

   A third relation comes from equating moduli: $|x+iy|^2 = |7+24i|$, i.e.

   $x^2+y^2 = \sqrt{7^2+24^2} = \sqrt{49+576} = \sqrt{625} = 25.$

   …

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6. CBSE 2023Set 2A2 marks
   Q.If $z_1 = -1$ and $z_2 = i$, then find $\mathrm{Arg}\left(\dfrac{z_1}{z_2}\right)$.

   ›Reveal solutionSolution

   Simplify $z_1/z_2$ to a single complex number first, then read off its argument.

   Given $z_1=-1$, $z_2=i$.

   $\frac{z_1}{z_2} = \frac{-1}{i} = \frac{-1}{i}\cdot\frac{-i}{-i} = \frac{i}{-i^2} = \frac{i}{1} = i,$

   using $i^2=-1$. So $z_1/z_2 = i = 0 + 1\cdot i$, which lies on the positive imaginary axis.

   …

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7. CBSE 2022Set 2A2 marks
   Q.Find the multiplicative inverse of $7 + 24i$.

   ›Reveal solutionSolution

   The inverse of $7+24i$ is $\dfrac{7}{625}-\dfrac{24}{625}i$.

   The multiplicative inverse of a non-zero complex number $z$ is $z^{-1}=\dfrac{1}{z}=\dfrac{\bar z}{z\bar z}=\dfrac{\bar z}{|z|^2}$.

   Here $z = 7+24i$, so $\bar z = 7-24i$ and $|z|^2 = 7^2+24^2 = 49+576 = 625$.

   …

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8. CBSE 2020Set 2A2 marks
   Q.Find the complex conjugate of $(3 + 4i)(2 - 3i)$.

   ›Reveal solutionSolution

   Multiply the two complex numbers using $i^2=-1$, then flip the sign of the imaginary part to get the conjugate.

   Expand the product:

   $(3+4i)(2-3i) = 3(2) + 3(-3i) + 4i(2) + 4i(-3i) = 6 - 9i + 8i - 12i^2$

   Since $i^2=-1$, we get $-12i^2 = 12$:

   $= 6 + 12 - 9i + 8i = 18 - i$

   …

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9. CBSE 2020Set 2A2 marks
   Q.Write $z = -\sqrt{3} + i$ in modulus-amplitude form.

   ›Reveal solutionSolution

   Find the modulus $r=|z|$ and the argument $\theta$ (measured from the positive real axis), then write $z=r(\cos\theta+i\sin\theta)$.

   Here $z = -\sqrt3 + i$, so $x=-\sqrt3$, $y=1$.

   Modulus:

   $r = \sqrt{x^2+y^2} = \sqrt{(-\sqrt3)^2+1^2} = \sqrt{3+1} = \sqrt4 = 2$

   Argument: since $x<0$ and $y>0$, the point lies in the second quadrant. The reference (acute) angle $\alpha$ satisfies

   $\tan\alpha = \left|\frac{y}{x}\right| = \frac{1}{\sqrt3} \implies \alpha = \frac{\pi}{6}$

   …

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10. CBSE 2019Set 2A2 marks
    Q.If $z = 2 - 3i$, then show that $z^2 - 4z + 13 = 0$.

    ›Reveal solutionSolution

    Direct substitution of $z=2-3i$ into $z^2-4z+13$ simplifies to $0$.

    Step 1 — Compute $z^2$.

    $z^2=(2-3i)^2 = 4 - 12i + 9i^2 = 4-12i-9 = -5-12i$ (using $i^2=-1$).

    Step 2 — Compute $4z$.

    $4z = 4(2-3i) = 8-12i$.

    Step 3 — Combine. …

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11. CBSE 2018Set 2A2 marks
    Q.Find the complex conjugate of $(2 + 5i)(-4 + 6i)$.

    ›Reveal solutionSolution

    Multiply the two complex numbers first, then flip the sign of the imaginary part to get the conjugate.

    First expand the product:

    $(2+5i)(-4+6i) = 2(-4) + 2(6i) + 5i(-4) + 5i(6i)$

    $= -8 + 12i - 20i + 30i^2$

    Since $i^2 = -1$, $30i^2 = -30$, so:

    $= -8 - 30 + (12i - 20i) = -38 - 8i$ …

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12. CBSE 2018Set 2A2 marks
    Q.If $x + iy = \operatorname{cis}\alpha \cdot \operatorname{cis}\beta$, then find the value of $x^2 + y^2$.

    ›Reveal solutionSolution

    $\operatorname{cis}\alpha\cdot\operatorname{cis}\beta$ is itself of the form $\cos\theta+i\sin\theta$, whose modulus is always $1$, so $x^2+y^2=1$.

    Recall $\operatorname{cis}\theta = \cos\theta + i\sin\theta$. Using the product rule for $\operatorname{cis}$ (which follows from the compound-angle formulas):

    $\operatorname{cis}\alpha\cdot\operatorname{cis}\beta = (\cos\alpha+i\sin\alpha)(\cos\beta+i\sin\beta) = \cos(\alpha+\beta) + i\sin(\alpha+\beta) = \operatorname{cis}(\alpha+\beta)$ …

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