How to put a complex number into the form a+ib if it has an exponent So I'm trying to put this into the form a+ib
$(1 + i)^{1000}$
(Hint: Use the polar form of the number). 
I know the polar form without the exponent would be
$√2(cos(π/4)+isin(π/4))$
Do i just throw the exponent on at the end? I'm not sure how this helps get it into a+ib form
 A: $$(1+i)^{1000}=((1+i)^2)^{500}=(2i)^{500}=(-4)^{250}=4^{250}+0i$$
A: The polar form that the hint is talking about is $re^{i\theta}$. Using this gives us $(re^{i\theta})^\alpha=r^\alpha e^{i\theta\alpha}$. In your case, $\theta=\pi/4$ so since $4|1000$ the complex part hoes away. This leaves us with $|1+i|^{1000}=\sqrt{2}^{1000}=2^{500}$.
A: As you figured, the polar form is $\left(\sqrt{2}(\cos{(\pi/4)} + i\sin{(\pi/4)})\right)^{1000} = 2^{500}\left(\mathrm{e}^{i\pi/4}\right)^{1000} = 2^{50}\mathrm{e}^{i250\pi} = 2^{500}\mathrm{e}^{i\cdot0} = 2^{500}$.
A: $$(1+i)^{1000}=e^{1000\bigr(\ln(1+i)\bigl)}$$
Use $$\ln(1+i)=\frac{\pi i}{4}+\frac{\ln(2)}{2}$$
$$(1+i)^{1000}=2^{500}e^{250\pi i}=2^{500}$$
A: Apropos of the answers from other posters applying DeMoivre's Theorem, the problem can be thought of in terms of the geometrical interpretation of multiplication between complex numbers.  When we multiply a number $ \ a + bi \ = \ \sqrt{a^2 + b^2}·cis(\theta) \ \ $ by a second number $ \ c + d \ = \ \sqrt{c^2 + d^2}·cis(\phi) \ \ , \  $ with $ \ \theta \ = \ \arctan\left(\frac{b}{a} \right) \ $ and $ \ \phi \ = \ \arctan\left(\frac{d}{c} \right) \ $ (adjusted to the appropriate quadrants), the product $$ \sqrt{a^2 + b^2} \ · \ \sqrt{c^2 + d^2} \ · \ cis(\theta + \phi) $$
can be regarded as the "re-scaling" (dilation) of the length $ \ \sqrt{a^2 + b^2} \ $ of a vector representing the first number by the factor $ \ \sqrt{c^2 + d^2} \ $ and a counter-clockwise rotation of the direction $ \ \theta \ $ of that vector by an angle $ \ \phi \ $ (or vice versa, of course, with the first number acting on the "vector" of the second number).
So for $ \ (1 + i)^2 \ = \ (1 + i)·(1 + i) \ \ , \ $ we can treat this as a vector of length $ \ \sqrt2 \ $ pointing in the direction $ \ \frac{\pi}{4} \ $ having its length multiplied by $ \ \sqrt2 \ $ and its direction rotated counter-clockwise by $ \ \frac{\pi}{4} \ \ , \ $ producing a "product vector" with length $ \ 2 \ $ and direction $ \ \frac{\pi}{2} \ \ , \ $ thus
$ \ 2  \ cis \left(\frac{\pi}{2} \right) \ \ $ or $ \ 0 + 2i \ \ . $  Repeated multiplication for $ \ (1 + i)^n \ $ creates a succession of vectors, the "tips" of which lie on an equiangular (or "logarithmic") spiral, with each new vector being $ \ \sqrt2 \ $ times longer than its predecessor and pointed  $ \ \frac{\pi}{4} \  $ radians counterclockwise relative to that predecessor.  We conclude that $ \ (1 + i)^{1000} \ \  $ corresponds to a vector with length of $ \ (\sqrt2)^{1000} \ = \ 2^{500} \ $ and direction $ \ \frac{1000·\pi}{4} \ = \ 250 \pi \ \equiv \ 0 \ \ , \ $ which is to say $ \ (1 + i)^{1000} \ = \ 2^{500} \ cis \ 0 \ = \ 2^{500} + 0·i \ \ . $
$$ \ \ $$
Alternatively, we can examine the implied "binomial-power".  We calculate $ \ ( 1 + 1 )^n \ $ by summing the terms $ \ \binom{n}{k}·1^k·1^{n - k} \ $ using the rows of the (Yang Hui/Pingala/Khayyam/Tartaglia/...) Pascal triangle,
$ 1 \quad \quad \quad \quad \quad \quad \quad \quad \quad \ = \ 1 \ = \ 2^0 $
$ 1 \ \ + \  \ 1 \quad \quad \quad \quad \quad \quad \ \ \ \ = \ 2 \ = \ 2^1 $
$ 1 \ \ + \  \ 2 \ \ + \ \ 1 \quad \quad \quad \quad \ \ = \ 4 \ = \ 2^2 $
$ 1 \ \ + \  \ 3 \ \ + \ \ 3 \ \ + \ \ 1 \quad \quad  = \ 8 \ = \ 2^3 \ \ , \  $ etc.
as a way to demonstrate that $ \ \sum_{k=0}^{n} \ \binom{n}{k} \ = \ 2^n \ \ . $  If we do something similar for the terms $ \ \binom{n}{k}·1^k·i^{n - k} \ $  of $ \ (1 + i)^n \ \ , \  $ we have (for the "even" rows)
$  \require{cancel} \cdots $
$ \cancel{1} \ \ + \  \ 2·i \ \ + \ \ \cancel{1·i^2} \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad   \ = \ 2i  $
$ 1 \ \ + \  \  \cancel{4·i} \ \ + \ \ 6·i^2 \ \ + \ \ \cancel{4·i^3} \ \ +  \ \ 1·i^4 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad   \ \  \ = \ -4 \ = \ (2i)^2   $
$ \cancel{1} \ \ + \  \ 6·i \ \ + \ \ \cancel{15·i^2} \ \ + \ \ 20·i^3 \ \ +  \ \ \cancel{15·i^4} \ \ +  \ \ 6·i^5 \ \ +  \ \ \cancel{1·i^6} \quad   = \ -8i \ = \ (2i)^3   $
$ 1 \ \ + \  \ \cancel{8·i} \ \ + \ \ 28·i^2 \ \ + \ \ \cancel{56·i^3} \ \ +  \ \ 70·i^4 \ \ + \ \ \cancel{56·i^5} \ \ + \ \ 28·i^6 \ \  +  \ \ \cancel{8·i^7} \ \ +  \ \ 1·i^8 \   = \ 16 \ = \ (2i)^4  \ \ , \ \text{etc.} $
So the binomial theorem establishes that $ \ (1 + i)^{2n} \ = \ (2i)^n \ \ , \ $ which Michael Rozenberg applies in his answer.
