Complex numbers - find real and imaginary parts of $z=(1+i)^{100}$ Currently studying for my calculus exam when I stumbled upon this example:
Find the real and imaginary parts of the following
$$z=(1+i)^{100}$$
Following the answer to this problem it's first stated that $1+i$ can be written in polar form as $\sqrt{2}e^{i\frac{\pi }{4}}$ which I totally get. Proceeding by rewriting it as:
$$z=\sqrt{2}^{100} * (e^{i\frac{\pi}{4}})^{100}$$
Then we get to the parts where I'm totally lost at the moment.
$$2^{50}*e^{i2\pi} = 2^{50}*e^{i\pi} = -2^{50}$$
Where the real part would be equal to $-2^{50}$ and the imaginary part to be equal to $0$ which I can see in the answer given.
However, the last line of simplification is what confuses me. Could someone explain what is done?
Thanks!
 A: What about if we start with $$(1+i)^2=1+2i-1=2i.$$ Then 
$$(1+i)^{100}=[(1+i)^2]^{50}=(2i)^{50}=2^{50}(i^2)^{25}=2^{50}(-1)^{25}=-2^{50}.$$
So the real part is $-2^{50}$ and the imaginary part is $0$.
A: It looks like there was a typo: 
$$\sqrt{2}^{100} \cdot \left(e^{\tfrac\pi4i}\right)^{100} = 2^{50} \cdot e^{25 \pi i} = 2^{50} \cdot e^{\pi i} = -2^{50}$$
The first equality is just multiplying powers, and they had $2\pi$ instead of $25\pi$. The second equality comes from the fact that $e^{2\pi i} = 1$. The third equality uses $e^{\pi i} = -1$
A: I believe the 'part where you are lost' should read:
$2^{50}∗e^{25*iπ}$ = $2^{50}∗e^{6*4iπ}∗e^{iπ}$ = $2^{50}∗e^{iπ}$ = $2^{-50}$
A: With a similar starting point to a given answer, but a different pathway:
$$(1+i)^2 = 2i$$
and squaring this
$$(1+i)^4 = -4 = - 2^2$$
and then
$$((1+i)^4)^{25} = (1+i)^{100} = (-)^{25} (2^{2\times25}) = -2^{50}$$
as before.
A: The squere root can be written as
\begin{equation}
\sqrt{x} = x^{\frac{1}{2}}
\end{equation}
So
\begin{equation}
\sqrt{2}^{100} = (2^{\frac{1}{2}})^{100}=2^{50}
\end{equation}
While
\begin{equation}
(e^{j \frac{\pi}{4}})^{100} = e^{j 25 \pi}=  e^{j \pi}  e^{j 24 \pi}=-1 e^{j 24 \pi} = -1
\end{equation}
