- Can someone explain to me, step by step, how to calculate this,
$$x=(-1-i)^{15}+(-1+i)^{11}$$
- Method 1 (Transform the numbers (−1−i) and (−1+i) to polar coordinates ) by Mr 5xum
\begin{align*} x&=(\sqrt{2})^{15}\exp(-\frac{45\pi}{4}i)+(\sqrt{2})^{11}\exp(\frac{33\pi}{4}i)\\ x&=128\sqrt{2}\exp(-\frac{45\pi}{4}i)+32\sqrt{2}\exp(\frac{33\pi}{4}i) \end{align*}
but i got stuck
- Method 2
note that $(1+i)^{2}=2i ,\quad (i-1)^{2}=-2i$ then :
\begin{align*}
x&=(-1)^{15}(1+i)^{15}+(i-1)^{11}\\
x&=-(1+i).(1+i)^{2.7}+(i-1)(i-1)^{2.5}\\
x&=-(1+i).(2i)^{7}+(i-1)(-2i)^{5}\\
x&=-(1+i).(-128i)+(i-1)(-32i)\\
x&=(-128+32)+(128+32)i\\
\end{align*}
this is easy way but what about the first method can someone explain to me, in details how to use it
Hint: Transform the numbers $(-1-i)$ and $(-1+i)$ to polar coordinates first. Then calculate their powers.