Hi,
steps appreciated in the question:
[squareroot(2) - i*squareroot(2)]^10
thanks
You need to apply De Moivre's Theorem
First, covert this to polar form:
You should get $\displaystyle z=2e^{i\left(\frac{3\pi}{4}\right)}$
Thus, $\displaystyle z^{10}=2^{10}e^{i\left(\frac{30\pi}{4}\right)}=2^{ 10}e^{i\left(-\frac{\pi}{2}+8\pi\right)}=2^{10}\left[\cos\!\left(-\frac{\pi}{2}\right)+i\sin\!\left(-\frac{\pi}{2}\right)\right]=-2^{10}i=\color{red}\boxed{-1024i}$
Does this make sense?