Considering $ (1+i)^n - (1 - i)^n $, Complex Analysis I have been working on problems from Complex Analysis by Ahlfors, and I got stuck in the following problem:
Evaluate:
$$
(1 + i)^n - (1-i)^n 
$$
I have just "reduced" to:
$$ 
(1 + i)^n - (1-i)^n  = \sum_{k=0} ^n i^k(1 - (-1)^k)
$$
by using expansion of each term.
Thanks.
 A: There are a number of spiffy techniques one could use on this problem, but Ahlfors doesn't get to conjugation and modulus until 1.3 and geometry of the complex plane until Section 2 (of Chapter 1), while this is still in 1.1. [I dug up my copy of the third edition to see how much was discussed to that point.]  
abel shows one approach in using the binomial theorem that lies within the (rather) limited means available.  Given what the author covers in this section, this is another possibility:
$$ ( 1 \ + \ i )^n \ + \ ( 1 \ - \ i )^n \ \ = \ \ ( 1 \ + \ i )^n \ \left[ \ 1 \  + \ \frac{( 1 \ - \ i )^n}{( 1 \ + \ i )^n} \ \right]  $$
$$ = \ \ ( 1 \ + \ i )^n \ \left[ \ 1 \  + \ \left(\frac{ 1 \ - \ i }{ 1 \ + \ i } \right)^n \ \right] \ \ = \ \ ( 1 \ + \ i )^n \ \left[ \ 1 \  + \ \left(\frac{ [ \ 1 \ - \ i \ ] \ [ \ 1 \ - \ i \ ]  }{ [ \ 1 \ + \ i \ ] [ \ 1 \ - \ i \ ] } 
\right)^n \ \right]  $$
[the conjugate is being applied as shown in that section, but Ahlfors hasn't called it that yet]
$$ = \ \ ( 1 \ + \ i )^n \ \left[ \ 1 \  + \ \left(\frac{  1 \ - \ 2i \ - \ 1     }{  2  } \right)^n \ \right] \ \ = \ \ ( 1 \ + \ i )^n \ [ \ 1 \  + \ ( -  i) ^n \ ] \ \ . $$
The binomial theorem can now be applied to the first factor:
$$ ( 1 \ + \ i )^n \ + \ ( 1 \ - \ i )^n $$ $$ = \ \ \left( \ 1 \ + \ \left( \begin{array}{c} n \\ 1 \end{array} \right) i \ + \ \left( \begin{array}{c} n \\ 2 \end{array} \right) i^2 \ + \ \ldots \ + \ \left( \begin{array}{c} n \\ n-1 \end{array} \right) i^{n-1} \ + \ i^n \ \right) \ [ \ 1 \  + \ ( - i) ^n \ ] \ \ .  $$
[The "typoed" version of the problem that David Cardozo originally posted is analogous:
$$ ( 1 \ + \ i )^n \ - \ ( 1 \ - \ i )^n $$ $$ = \ \ \left( \ 1 \ + \ \left( \begin{array}{c} n \\ 1 \end{array} \right) i \ + \ \left( \begin{array}{c} n \\ 2 \end{array} \right) i^2 \ + \ \ldots \ + \ \left( \begin{array}{c} n \\ n-1 \end{array} \right) i^{n-1} \ + \ i^n \ \right) \ [ \ 1 \  - \ ( - i) ^n \ ] \ \ . \ \ ] $$
$ \ \ $
Presumably, we'd like to consolidate this a bit. Because of that $ \ (-i)^n \ $ term in the second factor, that factor has a cycle of period 4. We see that this product is zero for $ \ n \ = \ 4m \ + \ 2 \ $ , with integer $ \ m \ \ge \ 0 \ $ .  [These will be "out-of-phase" with abel's expressions, since I am using Ahlfors' version of the problem.]  
For the other cases, we will write the first factor as
$$ \left[ \ 1 \ - \ \left( \begin{array}{c} n \\ 2 \end{array} \right) \ + \ \left( \begin{array}{c} n \\ 4 \end{array} \right) \ + \ \text{etc.} \ \right] \ \ + \ \ i \ \left[ \  \left( \begin{array}{c} n \\ 1 \end{array} \right) \ - \ \left( \begin{array}{c} n \\ 3 \end{array} \right) \ +  \ \left( \begin{array}{c} n \\ 5 \end{array} \right) \ - \ \text{etc.} \ \right] \ \ . $$
For $ \ n \ = \ 4m \ $ , the factor $ \ [ \ 1 \  + \ ( - i) ^n \ ] \ = \ 2 \ $ and the imaginary part of the binomial series is zero, owing to the symmetry of the binomial coefficients.  The real part also simplifies due to this symmetry, so we have
$$ ( 1 \ + \ i )^{4m} \ + \ ( 1 \ - \ i )^{4m} \ \ = \ \  \left[ \ 2 \cdot  1 \ - \ 2  \ \left( \begin{array}{c} 4m \\ 2 \end{array} \right) \ + \ 2 \ \left( \begin{array}{c} 4m \\ 4 \end{array} \right) \ - \ \ldots \ + \ \left( \begin{array}{c} 4m \\ 2m \end{array} \right) \ \right]  \cdot \ 2 \ \ . $$
The remaining cases are somewhat more complicated to work out:  for $ \ n \ = \ 4m \ + \ 1 \ $ and $ \ n \ = \ 4m \ + \ 3 \ $ , respectively, we obtain
$$ \left( \ \left[ \ 1 \ - \ \left( \begin{array}{c} n \\ 2 \end{array} \right) \ + \ \left( \begin{array}{c} n \\ 4 \end{array} \right) \ \ldots \ \pm \ n \ \right] \ \ + \ \ i \ \left[ \   n \ - \ \left( \begin{array}{c} n \\ 3 \end{array} \right) \ +  \ \left( \begin{array}{c} n \\ 5 \end{array} \right) \ \ldots \ \pm \ 1 \ \right] \ \right) \ \cdot \ ( 1 \ \mp \ i) \ \ . $$
For either of these cases, since $ \ n \ $ is odd, the number of binomial coefficients is even.  So the real part has the symmetry in which the first half of the terms are identical to the second half of them; also, the symmetry among the coefficients produces a sum which is always a power of 2 .  In the imaginary part, we do not get a simple alternation of signs (which we know gives a sum of zero for the binomial coefficients), but the "double-alternating" signs proves to have the same effect; the result is that the imaginary part is zero for these cases as well.
Hence, the expression $ \ ( 1 \ + \ i )^n \ + \ ( 1 \ - \ i )^n \ $ is always real; by analogous reasoning, the expression $ \ ( 1 \ + \ i )^n \ - \ ( 1 \ - \ i )^n \ $ is always pure imaginary.  We find the sequences (including the values  abel presents) for $ \ 0 \ \le \ n \ \le \ 9 \ $
$$ \ ( 1 \ + \ i )^n \ + \ ( 1 \ - \ i )^n \ \ : \ \ 2 \ , \ 2 \ , \ 0 \ , \ -4 \ , \ -8 \ , \ -8 \ , \ 0 \ , \ 16 \ , \ 32 \ , \ 32 \ \ \text{and} $$
$$ \ ( 1 \ + \ i )^n \ - \ ( 1 \ - \ i )^n \ \ : \ \ 0 \ , \ 2i \ , \ 4i \ , \ 4i \ , \ 0 \ , \ -8i \ , \ -16i \ , \ -16i \ , \ 0 \ , \ 32i \ \ . $$
[Incidentally, these results indicate the interesting identities$ ^* $
$$ ( 1 \ + \ i )^{4m} \ + \ ( 1 \ - \ i )^{4m} \ \ = \ \ ( 1 \ + \ i )^{4m+1} \ + \ ( 1 \ - \ i )^{4m+1} \ \ \text{and} $$ 
$$ ( 1 \ + \ i )^{4m+2} \ - \ ( 1 \ - \ i )^{4m+2} \ \ = \ \ ( 1 \ + \ i )^{4m+3} \ - \ ( 1 \ - \ i )^{4m+3} \ \ ] $$ 
$ ^* $ with unintentional alliteration on the theme of $ \ i \ $ ...
$ \ \ $
To be sure, this is a cumbersome description of the result, but it is a consequence of using Cartesian coordinates for the description of the complex values. Once you reach Section 2 and the use of polar coordinates, you will have the far more compact expressions
$$ \ ( 1 \ + \ i )^n \ + \ ( 1 \ - \ i )^n \ \ = \ \ 2^{(n+2)/2} \ \cos\left( \frac{n \pi}{4} \right) \ \ \text{and}  $$ $$( 1 \ + \ i )^n \ - \ ( 1 \ - \ i )^n \ \ = \ \ 2^{(n+2)/2} \ \sin\left( \frac{n \pi}{4} \right) \ i \ \ . $$
(The exponential factor simply grows by a factor of $ \ \sqrt{2} \ $ at each successive stage, but its product with the trigonometric factors create the complications in the sequences above.  The trigonometric factors also immediately explain the "out-of-phase" behavior between the two versions of the expression we have been evaluating.  These products follow from the methods being described by dustin and RikOsuave. )
A: ok. we can use the binomial theorem. we will check out for small values of $n.$
case $n = 1, \ z_1 = 1+ i -(1-i) = 2i$
case $n = 2, \ z_2 = (1+i)^2 - (1-i)^2 =(1 + 2i + i^2) -(1 - 2i + i^2) = 2(2i) = 4i $
case $n = 3, \ z_2 = (1+i)^3 - (1-i)^3 =(1 + 3i + 3i^2 + i^3) -(1 - 3i + 3i^2 - i^3) = 2(3i +i^3) = 4i$
case $n = 4, \ z_4 = (1 + i)^4 - (1 - i)^4 =2(4i + 4i^3) = 0$
case $n = 5, \ 
z_5 = (1 + i)^5 - (1 - i)^5 =2(5i + 10i^3 + i^5) = -8i $
case $n = 6, \ 
z_6 = (1 + i)^6 - (1 - i)^6 =2(6i + 20i^3 + 6i^5) = -16i$
case $n = 7, \ 
z_7 = (1 + i)^7 - (1 - i)^7 =2(7i + 35i^3 + 21i^5 + i^7) = -16i$
case $n = 8, \ 
z_8 = (1 + i)^8 - (1 - i)^8 =2(8i + 56i^3 + 56i^5 + 8i^7) = 0$
so it looks like the formula can be put in four classes: 
if $n$ is a multiple of $4,$ then $z_n = 0$ 
if $n = 4k+1,$ then $z_n = (2i)^{(n+1)/2}$
if $n = 4k+2, 4k+3,$ then $z_n = 2(2i)^{(n+1)/2}$
A: A complex number has both a modulus and angle. The modulus is the norm. Let $z=x+yi$. Then the modulus is
$$
\lvert z\rvert =\sqrt{x^2+y^2}
$$
and the angle is 
$$
\phi = \arctan(y/x).
$$
Now you can write the Complex number in exponential form
$$
z=\lvert z\rvert e^{i\phi}.
$$
You should be able to take it from here once you convert the complex numbers to exponential form. Remember to pay attention to the quadrant when you take the arctan.
A: Using the geometry of Complex numbers (which is the same as what was said in the comments, in different words), $(a+bi)^n$ , as multiplication of a number by itself n times. Then, when multiplying two complex numbers, you multiply their moduli and add the arguments. $1+i$ has length $(1+1)^{1/2}$ and has argument $\pi/4$ . Now you can put $1-i$ in the same form and see what happens when you raise the number to the $nth$ power.
