22 Most VSAQ’s of Complex Numbers Chapter in Inter 2nd Year Maths-2A (TS/AP)

Table of Contents

2 Marks

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 $$z = 2 – 3i$$

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

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

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

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