Question

If$a+ib=(1-\sqrt{3}i{)}^{100}$, then find $a+b$.

Answer 1

Given that

$\begin{array}{c}a+ib={\left(1-\sqrt{3i}\right)}^{100}\\ putr\cos \theta =1-----\left(i\right)\\ r\sin \theta =\sqrt{3}-----\left(ii\right)\end{array}$

Squaring and adding

$\begin{array}{c}{r}^2{\cos }^2\theta +r^2{\sin }^2\theta =4\\ \Rightarrow r^2\left({\cos }^2\theta +{\sin }^2\theta \right)=4\\ \Rightarrow r^2\times 1=4\\ \Rightarrow r=\sqrt{4}\\ \Rightarrow r=2\\ \tan \theta =\frac{r\sin \theta }{r\cos \theta }\\ \tan \theta =\frac{\sqrt{3}}{1}\\ \theta =\frac{\pi }{3}\\ \therefore a+ib={\left(r\cos \theta -ir\sin \theta \right)}^{100}\\ \Rightarrow a+ib=r^{100}{\left(\cos \theta -i\sin \right)}^{100}\\ \Rightarrow a+ib=r^{100}\left(\cos 100\theta -i\sin 100\theta \right)\\ \Rightarrow a+ib=2^{100}\left(\cos 100\theta -i\sin 100\theta \right)\end{array}$

Equating to real and imaginary part we get,

$\begin{array}{c}a=2^{100}\cos 100\theta \\ b=-2^{100}\sin 100\theta \end{array}$

$\therefore$

$\begin{array}{c}a+b=2^{100}\cos 100\theta -2^{100}\sin 100\theta \\ =2^{100}\left(\cos 100\theta -\sin 100\theta \right)\\ =2^{100}\left(\cos 100\frac{\pi }{3}-\sin 100\frac{\pi }{3}\right)\end{array}$