Find all the values of $(1+i)^{100} $

i used De Moivre's theorem to tackle this problem. $$(1+i)=\sqrt{2}(\cos(\pi/4)+i\sin(\pi/4)) $$ using the theorem, $$ (1+i)^{100}=\sqrt{2}(\cos(\pi/4)+i\sin(\pi/4))^{100}$$ after some calculations, $$=2^{50}(\cos(8n+1)25\pi+i\sin(8n+1)25\pi)$$ putting $n=0,1,2,\ldots,99$,we get the required values. My solution is correct ?

  • $\begingroup$ I think it should be just a complex number. Like if you put n = 1 into the formula then you get for the real part 2^50*cos(225pi) which is just 2^50*cos(pi) and cos(pi) is -1. $\endgroup$ – stuart stevenson Nov 23 '17 at 21:40
  • $\begingroup$ How do you mean find all the values of $(1+i)^{100}$? $\endgroup$ – kingW3 Nov 23 '17 at 21:42
  • $\begingroup$ I think op is confusing roots and integer powers. $\endgroup$ – randomgirl Nov 23 '17 at 21:43
  • $\begingroup$ you want to find all the values ? Does that expression has more than one value? $\endgroup$ – Guy Fsone Nov 23 '17 at 21:44
  • $\begingroup$ i did not come up with the question. This was a question in a test exam. i solved it like this. $\endgroup$ – user1157 Nov 23 '17 at 21:45

Notice that

$$2^{50}(\cos(8n+1)25\pi+i\sin(8n+1)25\pi)=2^{50}(\cos 25\pi+i\sin25\pi)$$

for all $n \in \mathbb{Z}$.

There is only one solution. Your $100$ solutions can be proven to be the same complex number using the property that sines and cosines are function with period $2\pi$.

If we square a complex number, we just get a single complex number rather than $2$.

If you are taking a square root, then perhaps you can find two solutions.


\begin{align}2^{50}(\cos(8n+1)25\pi+i\sin(8n+1)25\pi)&=2^{50}(\cos 25\pi+i\sin25\pi)\\&= 2^{50}(\cos \pi+i\sin\pi)\\ &= -2^{50}\end{align}

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  • $\begingroup$ so , $2^{50}$ and$-2^{50}$ ? $\endgroup$ – user1157 Nov 23 '17 at 21:48
  • $\begingroup$ $2^{50}=(-2)^{50}$ and $2^{50} \neq -2^{50}$. $\endgroup$ – Siong Thye Goh Nov 23 '17 at 21:49
  • $\begingroup$ Thanks a lot. So, my solution is correct. Even though i did not notice that they give the same solution. $\endgroup$ – user1157 Nov 23 '17 at 21:54

We have $(1+i)^{100} = (2i)^{50} = (-4)^{25} = - 2^{50} = -1125899906842624.$

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