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Evaluate $(1+i)^{11}$.

I got $(1+i)^{11} = 2^{11/2} [\cos (11/4π) + i\sin(11/4π)]$

I don't know how the angle change to $3/4\pi$

That is:
$$2^{11/2} [\cos(3/4π) + i\sin(3/4π)]$$

I think I have forgotten

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  • $\begingroup$ Hint: $\frac{11}{4}\pi=2 \pi + \frac{3}{4}\pi$ $\endgroup$ – dxiv Dec 9 '16 at 17:17
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$$(1+i)^{11}=(1+i)^{10}\times(1+i)$$ $$(2i)^5\times(1+i)=32i(1+i)=32(-1+i)$$

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  • $\begingroup$ So how did the angle become 3/4π? $\endgroup$ – Ashalley Samuel Dec 9 '16 at 17:51
  • $\begingroup$ There is no need to make the angle 3/4π. The method I have used is most convenient here. $\endgroup$ – Vidyanshu Mishra Dec 9 '16 at 17:53
  • $\begingroup$ So if i decide to use (1+i)^11 without breaking it into 32(-1+i) i will be correct with an angle of 11/4π right? That is using the de moive theorem $\endgroup$ – Ashalley Samuel Dec 9 '16 at 18:12
  • $\begingroup$ @AshalleySamuel, I can't understand what you are trying to say. $\endgroup$ – Vidyanshu Mishra Dec 9 '16 at 18:13
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$\cos(x)=\cos(x + 2\pi n) $ where $n\in\mathbb{Z}$, it is a periodic function, as is $\sin$. So a complex number with an argument of $11\pi/4$ is just the same as one with argument $3\pi/4$, you can change it by integer multiples of $2\pi$. The complex number lies in the second quadrant (top left)

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  • $\begingroup$ Oh thanks...i really appreciate. $\endgroup$ – Ashalley Samuel Dec 10 '16 at 0:04

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