Question

Express the following complex no in polar form.
(i) $(1-sin\alpha )+i\left(cos\alpha \right)$
(ii) $\frac{1-i}{cos\frac{\pi }{3}+isin\frac{\pi }{3}}=\frac{1-i}{\frac{1}{2}+i\frac{\sqrt{3}}{2}}$

Answer 1

(i) $(1-sin\alpha )+i\left(cos\alpha \right)$
$z=x+iy$
Plolar form, $z=r(cos\theta +isin\theta )$
$r=\sqrt{x^2+y^2}$
$tan\theta =\frac{y}{x}=\frac{cos\alpha }{1-sin\alpha }cos\theta =\frac{1-sin\alpha }{\sqrt{(cos\alpha {)}^2+(1-sin\alpha {)}^2}}sin\theta =\frac{cos\alpha }{\sqrt{(cos\alpha {)}^2+(1-sin\alpha {)}^2}}r=\sqrt{(1-sin\alpha {)}^2+(cos\alpha {)}^2}\therefore z=\sqrt{(1-sin\alpha {)}^2+(cos\alpha {)}^2}[\frac{1-sin\alpha }{\sqrt{(cos\alpha {)}^2}+(1-sin\alpha {)}^2}+i\frac{cos\alpha }{\sqrt{(cos\alpha {)}^2}+(1-sin\alpha {)}^2}]$.

(ii) $\frac{1-i}{cos\frac{\pi }{3}+isin\frac{\pi }{3}}=\frac{1-i}{\frac{1}{2}+i\frac{\sqrt{3}}{2}}$
$=\frac{2(1-i)(1-i\sqrt{3})}{(1+i\sqrt{3})(1-i\sqrt{3})}=\frac{2[1-i(\sqrt{3}+1)-\sqrt{3}]}{1+3}=\frac{1}{2}\left[\right(1-\sqrt{3})-i(\sqrt{3}+1\left)\right]r=\sqrt{(1-\sqrt{3}{)}^2+(\sqrt{3}+1{)}^2}=\sqrt{43-2\sqrt{3}+3+1+2\sqrt{3}}\Rightarrow r=2\sqrt{2}tan\theta =\frac{\sqrt{3}+1}{1-\sqrt{3}}=\frac{tan60+tan45}{1-\left(tan\theta 60\right)\left(tan45\right)}\Rightarrow tan\theta =tan\left(105\right)\Rightarrow \theta ={105}^{\circ }z=2\sqrt{2}(cos{105}^{\circ }+isin{105}^{\circ })$