# Using De Moivre's theorem . $(1+i)^{100}$

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 ?

• 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. – stuart stevenson Nov 23 '17 at 21:40
• How do you mean find all the values of $(1+i)^{100}$? – kingW3 Nov 23 '17 at 21:42
• I think op is confusing roots and integer powers. – randomgirl Nov 23 '17 at 21:43
• you want to find all the values ? Does that expression has more than one value? – Guy Fsone Nov 23 '17 at 21:44
• i did not come up with the question. This was a question in a test exam. i solved it like this. – user1157 Nov 23 '17 at 21:45

## 2 Answers

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.

Edit:

\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}

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

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