# Math Help - Complex numbers proof

1. ## Complex numbers proof

Hi i am struggling to get this proof to work.
(1+i)^n + (1-i)^n = 2^ (n/2)+1 * cos nPi/4
n being any natural number.
I have assuming i would have to convert the LHS into polar form r(cos(theta) + isin(theta)) to end up with a multiply on the RHS.
Any help in dealing with this question would be greatly appreciated. Thank you

2. Originally Posted by Webby
Hi i am struggling to get this proof to work.
(1+i)^n + (1-i)^n = 2^ (n/2)+1 * cos nPi/4
n being any natural number.
I have assuming i would have to convert the LHS into polar form r(cos(theta) + isin(theta)) to end up with a multiply on the RHS.
Any help in dealing with this question would be greatly appreciated. Thank you
You're on the right track to convert to polar form. You also need to use DeMoivre's Theorem.

$1 + i = \sqrt{2}\left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right)$
and
$1 - i = \sqrt{2}\left(\cos \frac{-\pi}{4} + i\sin \frac{-\pi}{4}\right)$

Using DeMoivre's Theorem:
\begin{aligned}
(1 + i)^n &= (\sqrt{2})^n \left(\cos \frac{n\pi}{4} + i\sin \frac{n\pi}{4}\right) \\
&= 2^{n/2}\left(\cos \frac{n\pi}{4} + i\sin \frac{n\pi}{4}\right)
\end{aligned}

\begin{aligned}
(1 - i)^n &= (\sqrt{2})^n \left(\cos \frac{-n\pi}{4} + i\sin \frac{-n\pi}{4}\right) \\
&= 2^{n/2}\left(\cos \frac{-n\pi}{4} + i\sin \frac{-n\pi}{4}\right)
\end{aligned}

Putting it all together:
\begin{aligned}
(1 + i)^n + (1 - i)^n &= 2^{n/2}\left(\cos \frac{n\pi}{4} + i\sin \frac{n\pi}{4}\right) + 2^{n/2}\left(\cos \frac{-n\pi}{4} + i\sin \frac{-n\pi}{4}\right) \\
&= 2^{n/2}\left(\cos \frac{n\pi}{4} + i\sin \frac{n\pi}{4} + \cos \frac{-n\pi}{4} + i\sin \frac{-n\pi}{4}\right) \\
&= 2^{n/2}\left[\left(\cos \frac{n\pi}{4} + \cos \frac{-n\pi}{4}\right) + i\left(\sin \frac{n\pi}{4} + \sin \frac{-n\pi}{4}\right)\right] \\
\end{aligned}

Since cosine is an even function, cos(-x) = cos x. That means that
$\cos \frac{-n\pi}{4} = \cos \frac{n\pi}{4}$.

And since sine is an odd function, sin(-x) = -sin x. That means that
$\sin \frac{-n\pi}{4} = -\sin \frac{n\pi}{4}$.

Replace:
$2^{n/2}\left[\left(\cos \frac{n\pi}{4} + \cos \frac{n\pi}{4}\right) + i\left(\sin \frac{n\pi}{4} - \sin \frac{n\pi}{4}\right)\right]$
\begin{aligned}
&= 2^{n/2}\left(2\cos \frac{n\pi}{4} + i(0)\right) \\
&= 2^{n/2}\left(2\cos \frac{n\pi}{4}\right) \\
&= 2^{n/2} \cdot 2\left(\cos \frac{n\pi}{4}\right) \\
&= 2^{n/2 + 1}\left(\cos \frac{n\pi}{4}\right)
\end{aligned}

01

3. Hello : this is the resolution