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

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.
