Question

Express the following complex numbers in the form $r(cos\theta +isin\theta )$.

$\left(i\right)1+itan\theta \left(ii\right)tan\alpha -i\left(iii\right)1-sin\alpha +icos\alpha \left(iv\right)\frac{1-i}{cos\frac{\pi }{3}+isin\frac{\pi }{3}}$

Answer 1

$\left(i\right)1+itan\alpha \text{Let}z=1+i\tan \alpha |z|=\sqrt{1+ta{n}^2\alpha }=sec\alpha (1+itan\alpha )=(1+i\frac{sin\alpha }{cos\alpha })=\left\{\frac{cos\alpha +isin\alpha }{cos\alpha }\right\}=sec\alpha (cos\alpha +isin\alpha )[0\le \alpha \le \frac{\pi }{2}]$

$\left(ii\right)tan\alpha -i\text{Let}z=tan\alpha -i|z|=\sqrt{ta{n}^2\alpha +(-1{)}^2}=sec\alpha arg\left(z\right)=ta{n}^{-1}\left(\frac{-1}{tan\alpha }\right)=ta{n}^{-1}(-cost\alpha )=ta{n}^{-1}\left(tan\alpha (\frac{\pi }{2}+\alpha )\right)=(\frac{\pi }{2}+\alpha )\text{Polar form is}:sec\alpha (cos(\frac{\pi }{2}+\alpha )+isin(\frac{\pi }{2}+\alpha ))$

$\left(iii\right)1-sin\alpha +icos\alpha \text{Let}z=1-sin\alpha +icos\alpha z=\sqrt{(1-sin\alpha {)}^2+co{s}^2\alpha }=\sqrt{1-2sin\alpha +si{n}^2\alpha +co{s}^2\alpha }=\sqrt{1-2sin\alpha +1}=\sqrt{2-2sin\alpha }=\sqrt{2(1-2sin\alpha )}=\sqrt{2(si{n}^2\frac{\alpha }{2}+co{s}^2\frac{\alpha }{2}2sin\frac{\alpha }{2}.cos\frac{\alpha }{2})}=\sqrt{2{(sin\frac{\alpha }{2}-cos\frac{\alpha }{2})}^2}=\sqrt{2}(sin\frac{\alpha }{2}-cos\frac{\alpha }{2})\text{Polar Form of z},z=\sqrt{2}(sin\frac{\alpha }{2}-cos\frac{\alpha }{2})\{cos(\frac{\pi }{4}+\frac{\alpha }{2})+isin(\frac{\pi }{4}+\frac{\alpha }{2})\}[0\le \alpha \le \frac{\alpha }{2}]$

$\left(iv\right)\frac{1-i}{cos\frac{\pi }{3}+isin\frac{\pi }{3}}\text{Let}z=\frac{1-i}{cos\frac{\pi }{3}+isin\frac{\pi }{3}}=\frac{1-i}{\frac{1}{2}+i\frac{\sqrt{3}}{2}}=\frac{2(1-i)}{(1+i-\sqrt{3})}=\frac{2(1-i)(1-i\sqrt{3})}{(1+i\sqrt{3})(1-i\sqrt{3})}=\frac{2(1-i)(1-i\sqrt{3})}{(1^2-(i\sqrt{3}{)}^2)}=\frac{2(1-i\sqrt{3}-i-\sqrt{3})}{1+3}=\frac{2\left(\right(1-\sqrt{3})-i(1+\sqrt{3}\left)\right)}{1+3}=\frac{(1-\sqrt{3})-i(1+\sqrt{3})}{2}|z|=\left|\sqrt{{\left(\frac{1-\sqrt{3}}{2}\right)}^2+{\left(\frac{1+\sqrt{3}}{2}\right)}^2}\right|=2arg\left(z\right)=ta{n}^{-1}\left[\frac{\frac{(1+\sqrt{3})}{2}}{\frac{(1-\sqrt{3})}{2}}\right]=ta{n}^{-1}\left(\frac{1+\sqrt{3}}{1-\sqrt{3}}\right)=ta{n}^{-1}\left(\frac{(1+\sqrt{3})(1+\sqrt{3})}{(1-\sqrt{3})(1-\sqrt{3})}\right)=ta{n}^{-1}\left(\frac{1+3+2\sqrt{3}}{1-3}\right)=ta{n}^{-1}\left(\frac{4+2\sqrt{3}}{-2}\right)=ta{n}^{-1}(-(2+\sqrt{3}{)}^2)$