Write $(1+i )$ in polar form, then use this to compute $(1+i)^{10}$ I understand how to write $(1+i)$ in polar form, but how do I use it to compute $(1+i)^{10}$?
Thanks for the help!
 A: $$1+i=|1+i|e^{\arg(1+i)i}=$$
$$\sqrt{\Re(1+i)^2+\Im(1+i)^2}e^{\arctan\left(\frac{\Im(1+i)}{\Re(1+i)}\right)i}=$$
$$\sqrt{1^2+1^2}e^{\arctan\left(\frac{1}{1}\right)i}=$$
$$\sqrt{2}e^{\arctan\left(1\right)i}=$$
$$\sqrt{2}e^{\frac{\pi}{4}i}$$
So:
$$\left(1+i\right)^{10}=$$
$$\left(\sqrt{2}e^{\frac{\pi}{4}i}\right)^{10}=$$
$$\left(\sqrt{2}\right)^{10}\cdot e^{10\cdot\frac{\pi}{4}i}=$$
$$\sqrt{2^{10}}e^{\frac{10\pi}{4}i}=$$
$$\sqrt{1024}e^{\frac{5\pi}{2}i}=$$
$$32e^{\left(2\pi+\frac{\pi}{2}\right)i}=$$
$$32e^{\frac{\pi}{2}i}=$$
$$32i$$
A: Polar form is $r(\cos\theta + i\sin\theta)$.  The $10$th power is
$$
\Big(r(\cos\theta + i\sin\theta)\Big)^{10} = r^{10}(\cos(10\,\theta) + i\sin(10\,\theta)).
$$
The fact that
$$
(\cos\theta + i\sin\theta)^n = \cos(n\theta)+i\sin(n\theta)
$$
appears to be the part you missed.  It is sometimes called de Moivre's formula, after Abraham de Moivre, an 18th-century mathematician. I was puzzled by its being called a theorem when I first saw it in a secondary-school textbook, because I had been taught some years earlier that this is the definition of multiplication of complex numbers.
A: Hint: $a+bi=re^{i\theta}=r(\cos \theta +i \sin \theta)$ where $r=\sqrt{a^2+b^2}$ and $\tan \theta = \frac{b}{a}$.
$(a+bi)^k=r^k e^{i k \theta}=r^k(\cos (k \theta) +i \sin (k\theta))$
