- Thread starter
- #1
Petrus
Well-known member
- Feb 21, 2013
- 739
Hello MHB,
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,
\(\displaystyle |\rangle\)

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,
\(\displaystyle |\rangle\)