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
The polar form that the hint is talking about is $re^$. Using this gives us $(re^)^\alpha=r^\alpha e^$. In your case, $\theta=\pi/4$ so since $4|1000$ the complex part hoes away. This leaves us with $|1+i|^=\sqrt^=2^$.
answered Nov 21, 2017 at 2:45 Stella Biderman Stella Biderman 31.2k 6 6 gold badges 47 47 silver badges 93 93 bronze badges $\begingroup$ answered Nov 21, 2017 at 2:47 Abhiram Natarajan Abhiram Natarajan 1,226 7 7 silver badges 21 21 bronze badges $\begingroup$ answered Nov 21, 2017 at 2:48 4,027 1 1 gold badge 11 11 silver badges 17 17 bronze badges $\begingroup$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> \ $ having its length multiplied by $ \ \sqrt2 \ $ and its direction rotated counter-clockwise by $ \ \frac<\pi> \ \ , \ $ producing a "product vector" with length $ \ 2 \ $ and direction $ \ \frac<\pi> \ \ , \ $ thus $ \ 2 \ cis \left(\frac<\pi> \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> \ $ radians counterclockwise relative to that predecessor. We conclude that $ \ (1 + i)^ \ \ $ corresponds to a vector with length of $ \ (\sqrt2)^ \ = \ 2^ \ $ and direction $ \ \frac<1000·\pi> \ = \ 250 \pi \ \equiv \ 0 \ \ , \ $ which is to say $ \ (1 + i)^ \ = \ 2^ \ cis \ 0 \ = \ 2^ + 0·i \ \ . $
Alternatively, we can examine the implied "binomial-power". We calculate $ \ ( 1 + 1 )^n \ $ by summing the terms $ \ \binom·1^k·1^ \ $ 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_^ \ \binom \ = \ 2^n \ \ . $ If we do something similar for the terms $ \ \binom·1^k·i^ \ $ of $ \ (1 + i)^n \ \ , \ $ we have (for the "even" rows)
$ \cancel \ \ + \ \ 2·i \ \ + \ \ \cancel \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ = \ 2i $
$ 1 \ \ + \ \ \cancel \ \ + \ \ 6·i^2 \ \ + \ \ \cancel \ \ + \ \ 1·i^4 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \ \ = \ -4 \ = \ (2i)^2 $
$ \cancel \ \ + \ \ 6·i \ \ + \ \ \cancel \ \ + \ \ 20·i^3 \ \ + \ \ \cancel \ \ + \ \ 6·i^5 \ \ + \ \ \cancel \quad = \ -8i \ = \ (2i)^3 $
$ 1 \ \ + \ \ \cancel \ \ + \ \ 28·i^2 \ \ + \ \ \cancel \ \ + \ \ 70·i^4 \ \ + \ \ \cancel \ \ + \ \ 28·i^6 \ \ + \ \ \cancel \ \ + \ \ 1·i^8 \ = \ 16 \ = \ (2i)^4 \ \ , \ \text $
So the binomial theorem establishes that $ \ (1 + i)^ \ = \ (2i)^n \ \ , \ $ which Michael Rozenberg applies in his answer.