Writing $(1+i)^{18}$ in the form $a+bi$ How do I write 
$ (1+i)^{18}$ in the form $a+bi$? It is not a good method to simply expand it? Is there something else I can employ? 
 A: Note that the complex number $1+i$ can be written as $\sqrt{2}\left(\cos \frac{\pi}{4} + i \sin  \frac{\pi}{4} \right)$. This is called the "polar form" of a complex number, as opposed to the "rectangular" or cartesian representation, which has the form $a+ib$.
Khan Academy summarizes the polar form of a complex number as follows.

The polar form of complex numbers emphasizes their graphical attributes: the absolute value (the distance of the number from the origin in the complex plane) and the angle (the angle that the number forms with the positive Real axis). These are also called the modulus and argument, respectively.

Once we have converted a complex number into a polar representation, we can easily apply De Moivre's Theorem to raise it to an arbitrary exponent:
$$\left[r\left(\cos x + i \sin x\right)\right]^n = r^n \left(\cos (nx) + i \sin (nx)\right)$$
In your case,
\begin{align}
(1+i)^{18} &= \left[\sqrt{2}\left(\cos \frac{\pi}{4} + i \sin  \frac{\pi}{4} \right)\right]^{18} =2^{18/2} \left(\cos \frac{18\pi}{4} + i \sin  \frac{18\pi}{4} \right)\\
 &= 512\left( \cos \frac{\pi}{2} + i\sin \frac{\pi}{2} \right) = 512(0+i) =512i
\end{align}
A: I would say
$(1+i)^{18}=((1+i)^2)^9=(2i)^9=512i^9=512i$
A: To express $z=1+i$ in the form of $r(\cos\theta+i\sin\theta)$, we need to calculate the absolute value $r$ and argument $\theta$
$$\mbox{Absolute value: }r=\sqrt{1^2+1^2}=\sqrt{2}$$
$$\mbox{Argument: }\theta=\arctan\dfrac11=\dfrac{\pi}{4}$$
Now, by using DeMoiver's theorem, we get
$$z^{18}=\left[\sqrt{2}\left(\cos\dfrac{\pi}{4}+\sin\dfrac{\pi}{4}\right)\right]^{18}$$
$$=2^{18/2} \left(\cos \frac{18\pi}{4} + i \sin  \frac{18\pi}{4} \right)$$
$$=512\left(\cos\dfrac{\pi}{2}+i\sin\dfrac{\pi}{2}\right)$$
$$=512i$$
Therefore, $(1+i)^{18}=512i$
