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#### Petrus

##### Well-known member

- Feb 21, 2013

- 739

I got no clue about that, well I start wounder but I don't think so but is that same as

\(\displaystyle (\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}})^{56}\)

then I could go to polar form and get

\(\displaystyle r=1\) that means \(\displaystyle cos\theta=\frac{1}{\sqrt{2}} \ sin\theta=\frac{1}{\sqrt{2}}\) that means we got \(\displaystyle (e^{i\frac{\pi}{4}})^{56} <=>e^{i14\pi} \) so we got \(\displaystyle 1^{56}(1+0i) =1\)

Is this correct?

Regards,

\(\displaystyle |\rangle\)