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.
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:
$$ ( 1 \ + \ i )^n \ - \ ( 1 \ - \ i )^n $$ $$ = \ \ \left( \ 1 \ + \ \left( \begin{array}{c} n \\ 1 \end{array} \right) i \ + \ \left( \begin{array}{c} n \\ 2 \end{array} \right) i^2 \ + \ \ldots \ + \ \left( \begin{array}{c} n \\ n-1 \end{array} \right) i^{n-1} \ + \ i^n \ \right) \ [ \ 1 \ - \ ( - i) ^n \ ] \ \ . \ \ ] $$
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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. )
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 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.
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.
