Complex Numbers (VSAQs)

Maths-2A | 1. Complex Numbers – VSAQs:
Welcome to VSAQs in Chapter 1: Complex Numbers. This page contains the most important VSAQs in this chapter. Aim to secure top marks in your exams by understanding these clear and straightforward Very Short Answer Questions.

Overview:

VSAQ-1 : Write the complex number (2-3i) (3+4i) in the form A+iB

$$(2-3i)(3+4i) = 2(3) + 2(4i) – 3i(3) – 3i(4i)$$

$$= 6 + 8i – 9i – 12i^2$$

$$= 6 + 8i – 9i + 12$$

$$= 18 – i$$

VSAQ-2 : Express (1 – i)³(1 + i) in the form of a + ib

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

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

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

$$= (-2i)(2) = -4i = 0+i(-4)$$

VSAQ-3 : Write the conjugate of (3 + 4i)(2 – 3i)

$$(3+4i)(2-3i)$$

$$= 3(2) – 3(3i) + 4i(2) – 4i(3i)$$

$$= 6 – 9i + 8i – 12i^2$$

Since i²=−1, we can substitute and continue:

$$= 6 – 9i + 8i + 12$$

$$= 18 – i$$

VSAQ-4 : Write the conjugate of (2 + 5i)(-4 + 6i)

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

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

$$= -8 + 12i – 20i + 30i^2$$

Since i²=−1, we substitute and simplify:

$$= -8 + 12i – 20i – 30$$

$$= -38 – 8i$$

VSAQ-5 : Write the complex number (a-ib)/(a+ib) in the form A + ib

$$\frac{a-ib}{a+ib} = \frac{(a-ib)(a-ib)}{(a+ib)(a-ib)} = \frac{(a-ib)^2}{a^2+b^2}$$

$$= \frac{a^2 + i^2b^2 – 2abi}{a^2 + b^2} = \frac{a^2 – b^2 – 2abi}{a^2 + b^2}$$

$$= \frac{a^2 – b^2}{a^2 + b^2} + \frac{-2ab}{a^2 + b^2}i$$

VSAQ-6 : Find the real and imaginary parts of the complex number (a+ib)/(a-ib)

$$\frac{a + ib}{a – ib} = \frac{(a + ib)(a + ib)}{(a – ib)(a + ib)} = \frac{(a + ib)^2}{a^2 + b^2}$$

$$= \frac{a^2 + 2abi – b^2}{a^2 + b^2} = \frac{a^2 – b^2}{a^2 + b^2} + \frac{2ab}{a^2 + b^2}i$$

$$\frac{a^2 – b^2}{a^2 + b^2} + \frac{2ab}{a^2 + b^2}i$$

The real part (R.P) is $$\frac{a^2 – b^2}{a^2 + b^2}$$

The imaginary part (I.P) is $$\frac{2ab}{a^2 + b^2}$$

VSAQ-7 : Find the multiplicative inverse of 7 + 24 i

The multiplicative inverse of 7+24i:

$$\frac{1}{7 + 24i} = \frac{7 – 24i}{(7 + 24i)(7 – 24i)} = \frac{7 – 24i}{7^2 + (24)^2}$$

$$= \frac{7 – 24i}{49 + 576} = \frac{7 – 24i}{625}$$

$$= \frac{7}{625} – \frac{24i}{625}$$

VSAQ- 8 : Find the multiplicative inverse of √5+3i

The multiplicative inverse of $$\sqrt{5} + 3i$$ is

$$\frac{1}{\sqrt{5} + 3i} = \frac{\sqrt{5} – 3i}{(\sqrt{5} + 3i)(\sqrt{5} – 3i)} = \frac{\sqrt{5} – 3i}{5 + 9}$$

$$= \frac{\sqrt{5} – 3i}{14} = \frac{\sqrt{5}}{14} – \frac{3i}{14}$$

VSAQ-9 : Find the square root of 7+24i

Let $$7 + 24i = a + bi$$ where $$a = 7$$ $$b = 24$$

$$r = \sqrt{a^2 + b^2} = \sqrt{7^2 + 24^2}$$

$$r = \sqrt{49 + 576} = \sqrt{625} = 25$$

$$\sqrt{a + bi} = \pm \left(\sqrt{\frac{r + a}{2}} + i\sqrt{\frac{r – a}{2}}\right)$$

$$\sqrt{7 + 24i} = \pm \left(\sqrt{\frac{25 + 7}{2}} + i\sqrt{\frac{25 – 7}{2}}\right)$$

$$= \pm \left(\sqrt{\frac{32}{2}} + i\sqrt{\frac{18}{2}}\right) = \pm \left(\sqrt{16} + i\sqrt{9}\right)$$

$$= \pm (4 + 3i)$$

VSAQ-10 : Find the square root of -5 + 12i

Given: $$-5 + 12i = a + bi$$ where $$a = -5$$ $$b = 12$$

$$r = \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$$

$$\sqrt{a + ib} = \pm \left( \sqrt{\frac{r+a}{2}} + i\sqrt{\frac{r-a}{2}} \right)$$

$$\sqrt{-5 + 12i} = \pm \left( \sqrt{\frac{13 – (-5)}{2}} + i\sqrt{\frac{13 – 5}{2}} \right)$$

$$\sqrt{-5 + 12i} = \pm \left( \sqrt{\frac{13 + 5}{2}} + i\sqrt{\frac{13 – (-5)}{2}} \right)$$

$$\sqrt{-5 + 12i} = \pm(2 + 3i)$$

VSAQ-11 : Write z = -√7+i√21 in the polar form

Given the equation $$-\sqrt{7} + i\sqrt{21} = x + iy$$

$$x = -\sqrt{7}, \quad y = \sqrt{21}$$

$$r = \sqrt{x^2 + y^2} = \sqrt{(-\sqrt{7})^2 + (\sqrt{21})^2}$$

$$= \sqrt{7 + 21} = \sqrt{28} = \sqrt{4 \times 7} = 2\sqrt{7}$$

$$tan\theta = \frac{y}{x} \Rightarrow tan\theta = \frac{\sqrt{21}}{-\sqrt{7}} = -\sqrt{3} = tan\frac{2\pi}{3}$$

$$\Rightarrow \theta = \frac{2\pi}{3}$$

The polar form of $$-\sqrt{7} + i\sqrt{21}$$

$$r(cos\theta + isin\theta) = 2\sqrt{7}(cos\frac{2\pi}{3} + isin\frac{2\pi}{3})$$

VSAQ-12 : Express 1 + i√3 in the modulus-amplitude form

Given the equation $$1 + i\sqrt{3} = x + iy$$

$$x = 1, \quad y = \sqrt{3}$$

$$r = \sqrt{x^2 + y^2} = \sqrt{(1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2$$

$$tan\theta = \frac{y}{x} \Rightarrow tan\theta = \frac{\sqrt{3}}{1} = \sqrt{3} = tan\frac{\pi}{3}$$

$$\Rightarrow \theta = \frac{\pi}{3}$$

The polar form of $$1 + i\sqrt{3}$$

$$r(\cos\theta + i\sin\theta) = 2(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3})$$

VSAQ-13 : Express -1 –i in the modulus amplitude form

Given the equation $$-1 – i = x + iy$$

$$x = -1, \quad y = -1$$

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

$$tan\theta = \frac{y}{x} \Rightarrow tan\theta = \frac{-1}{-1} = 1 = tan\left(\frac{\pi}{4}\right)$$

$$\theta = \frac{3\pi}{4}$$

The polar form is: $$r(\cos\theta + i\sin\theta) = \sqrt{2}(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4})$$

VSAQ-14 : Express 1- i in the modulus amplitude form

Given the equation $$1 – i = x + iy$$

$$x = 1, \quad y = -1$$

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

$$tan\theta = \frac{y}{x} \Rightarrow tan\theta = \frac{-1}{1} = -1 = tan\left(-\frac{\pi}{4}\right)$$

$$\theta = -\frac{\pi}{4}$$

The polar form is:

$$r(\cos\theta + i\sin\theta) = \sqrt{2}(\cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4}))$$

$$= \sqrt{2}(\cos\frac{\pi}{4} – i\sin\frac{\pi}{4})$$

VSAQ-15 : Express the complex number -√3+i in the modulus amplitude form

Given the equation $$-\sqrt{3} + i = x + iy$$

$$x = -\sqrt{3}, \quad y = 1$$

$$r = \sqrt{x^2 + y^2} = \sqrt{(-\sqrt{3})^2 + (1)^2} = \sqrt{3 + 1} = \sqrt{4} = 2$$

To find θ, we calculate $$\tan^{-1}\left(\frac{y}{x}\right)$$

$$\tan\theta = \frac{y}{x} = \frac{1}{-\sqrt{3}} = -\frac{1}{\sqrt{3}}$$

Therefore, the modulus-amplitude (polar) form is:

$$r(\cos\theta + i\sin\theta) = 2(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6})$$

VSAQ-16 : Express -1 -i√3 in the mod-amp form

Given: $$-1 – i\sqrt{3} = x + iy \Rightarrow x = -1, y = -\sqrt{3}$$

$$r = \sqrt{x^2 + y^2} = \sqrt{(-1)^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2$$

$$\theta = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}\left(\frac{-\sqrt{3}}{-1}\right) = \tan^{-1}(\sqrt{3}) = -\frac{2\pi}{3}$$

$$r(\cos\theta + i\sin\theta) = 2\left(\cos\left(-\frac{2\pi}{3}\right) + i\sin\left(-\frac{2\pi}{3}\right)\right)$$

$$= 2\left(\cos\frac{2\pi}{3} – i\sin\frac{2\pi}{3}\right)$$

VSAQ-17 : If z₁ = -1, z₂ = -i, then find Arg(z₁z₂)

$$\text{Arg}(-1) = \pi$$

$$\text{Arg}(-i) = -\frac{\pi}{2}$$

$$\text{Arg}(z_1z_2) = \text{Arg}(z_1) + \text{Arg}(z_2)$$

$$\text{Arg}(-1 \cdot -i) = \text{Arg}(-1) + \text{Arg}(-i) = \pi – \frac{\pi}{2} = \frac{\pi}{2}$$

VSAQ-18 : If z₁ = -1, Z₂ = i, find Arg(z₁/z₂)

We know that: $$\text{Arg}(-1) = \pi$$ $$\text{Arg}(i) = \frac{\pi}{2}$$

$$\text{Arg}\left(\frac{z_1}{z_2}\right) = \text{Arg}(z_1) – \text{Arg}(z_2)$$

$$\text{Arg}\left(\frac{-1}{i}\right) = \text{Arg}(-1) – \text{Arg}(i) = \pi – \frac{\pi}{2} = \frac{\pi}{2}$$

VSAQ-19 : If Arg (z₁) , Arg z₂are π∕5,π∕3 then find Arg(z₁)+Arg(z₂)

Given that $$\text{Arg}(z_1) = \pi/5 \Rightarrow \text{Arg}(z_1) = -\pi/5$$

Also, $$\text{Arg}(z_2) = \pi/3$$

$$\text{Arg}(z_1) + \text{Arg}(z_2) = \pi/5 + \pi/3 = \pi/2 – \pi/5 = 5\pi-3\pi/15 = 2\pi/15$$

VSAQ-20 : If (a+ib)² = x+iy find (x²+y²)

$$(a+ib)^2 = x + iy \Rightarrow (a+ib)(a+ib) = x + iy$$

$$\Rightarrow |a+ib| |a+ib| = |x+iy|$$

$$\Rightarrow \sqrt{a^2 + b^2} . \sqrt{a^2 + b^2} = \sqrt{x^2 + y^2}$$

$$\Rightarrow a^2 + b^2 = \sqrt{x^2 + y^2}$$

Squaring on both sides, we get $$(a^2 + b^2)^2 = x^2 + y^2$$

VSAQ-21 : If x + iy = cisα.cisβ, then find the value of x²+y²

Given that $$x+iy = \text{cis}\alpha \cdot \text{cis}\beta = \text{cis}(\alpha+\beta) = \cos(\alpha+\beta)+i\sin(\alpha+\beta)$$

Equating the real parts, we get $$x = \cos(\alpha+\beta)$$

Equating the imaginary parts, we get $$y = \sin(\alpha+\beta)$$

$$x^2 + y^2 = \cos^2(\alpha+\beta) + \sin^2(\alpha+\beta) = 1$$

VSAQ-22 : If z = 2 – 3i, show that z² – 4z + 13 = 0

Given that $$z = 2 – 3i$$

$$\Rightarrow z-2 = -3i$$

$$\Rightarrow (z-2)^2 = 9i^2$$

$$\Rightarrow z^2-4z+4 = -9$$

$$\Rightarrow z^2-4z+13=0$$