Question

Convert 1+3i÷1-2i into polar form

Answer 1

Answer:-

$\red{\bigstar}$ Polar form $\large\leadsto\boxed{\rm\purple{\sqrt{2} \bigg(cos \dfrac{3\pi}{4} + i \: sin\dfrac{3 \pi}{4}\bigg)}}$

• Given:-

$\sf \dfrac{1 + 3i}{1 - 2i}$

• Solution:-

$\sf z = \dfrac{1 + 3i}{1 - 2i}$

Rationalising:-

→ $\sf \dfrac{1+3i}{1-2i} \times \dfrac{1+2i}{1+2i}$

→ $\sf \dfrac{1+2i+3i-6}{1+4}$

→ $\sf \dfrac{-5 + 5i}{5}$

→ $\sf \dfrac{5(-1 + i)}{5}$

→ $\bf -1 + i$

Let r cosθ = -1 , r sinθ = 1

• Squaring both sides and adding:-

→ $\sf r^2(cos^{2} \theta + sin^{2} \theta = 1+1$

→ $\sf r^2(cos^{2} \theta + sin^{2} \theta = 2$

→ $\sf r^2 = 2 \dashrightarrow\bf\red{[\because cos^2 \theta + sin^2 \theta = 1]}$

→ $\sf r = \sqrt{2}\dashrightarrow\bf\red{[r>0]}$

$\therefore$

★ $\sf \sqrt{2} cos \theta = -1$

★ $\sf \sqrt{2} sin \theta = 1$

→ $\sf cos \theta = \dfrac{-1}{\sqrt{2}}$

→ $\sf sin \theta = \dfrac{1}{\sqrt{2}}$

$\therefore$

$\sf \theta = \pi - \dfrac{\pi}{4} = \dfrac{3\pi}{4}\dashrightarrow\bf\red{[\because \theta \: lies \: in \: 2nd \: Quadrant]}$

→ $\sf z = r \: cos \theta + i \: r \: sin \theta$

→ $\sf \sqrt{2} cos \dfrac{3\pi}{4} + i \sqrt{2} sin \dfrac{3\pi}{4}$

→ $\sf \sqrt{2} \bigg(cos \dfrac{3\pi}{4} + i \: sin\dfrac{3 \pi}{4}\bigg)$

★ Hence, it is the polar form.