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The result of summing a conjugate pair of numbers each raised to the power $n$:

$$ (a + bi)^n + (a - bi)^n $$

Produces a real number where $a + bi$ is a complex number.

Given the result is real, is there a simplified way to express the above expression in terms of $a$ and $b$ involving no imaginary number $i$ in the simplified expression?

keith
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  • Hint: binomial theorem + treat odd and even separately – b00n heT May 25 '16 at 11:04
  • @b00n heT, would that be a simplification or a much much larger expression of sums (if $n$ is large)? – keith May 25 '16 at 11:10
  • Well... it depends on what you refer to as "simple"; another way to "simplify" it, would be to use polar coordinates, as both summands share the same modulus – b00n heT May 25 '16 at 11:15

4 Answers4

2

Let $z=|z|(cos (\alpha) + isin(\alpha))$ therefore $z^n=|z|^n(cos (n \alpha) + isin (n\alpha))$ so the expression becomes $2|z|^ncos (n\alpha)$. We know $|z|=\sqrt {a^2+b^2}$, also $\alpha$ can be expressed in terms of $a,b$ (using cotangent).

2

Well, \begin{align} a+bi &= re^{i \theta}, \qquad (r=|z|) \\ a-bi &= re^{-i \theta} \end{align} Thus. \begin{align} (a+bi)^{n}+(a-bi)^{n} &= \left(re^{i \theta} \right)^{n}+ \left(re^{-i \theta}\right)^{n} \\ &= r^{n}\left( e^{in \theta}+e^{-in \theta}\right) \\ &= 2|z|^{n} \cos n \theta \end{align}

0

By De Moivre's Formula we have, for $a+bi=r(\cos \theta+i\sin \theta)$ $$(a+bi)^n+(a-bi)^n=\\r^n(\cos n\theta+i\sin n\theta)+r^n(\cos (-n\theta)+i\sin (-n\theta))=\\2r^n\cos n\theta$$


Since $\cos n\theta=\cos (-n\theta)$ and $\sin n\theta=-\sin(-n\theta)$

Now we use the fact that $$r=\sqrt{a^2+b^2}$$ and $$L(a,b)=\cos n\theta=\begin{cases} \cos\theta\Big[1-\frac{(n^2-1^2)\sin^2\theta}{2!}+\frac{(n^2-1^2)(n^2-3^2)\sin^2\theta^4}{4!}-\cdots\Big] \text{n odd}\\1-\frac{n^2\sin^2\theta}{2!}+\frac{n^2(n^2-2^2)\sin^4\theta}{4!}-\cdots \text{n even} \end{cases}$$

where $$\cos\theta=\frac{a}{\sqrt{a^2+b^2}},\sin\theta=\frac{b} {\sqrt{a^2+b^2}}$$

to obtain the expression $$S(a,b)=2(\sqrt{a^2+b^2})^nL(a,b)$$ which is not dependant on $i$.

MathematicianByMistake
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0

Let's let $ S = (a + bi)^n + (a - bi)^n $. We can rewrite this sum in terms of binomial coefficients as follows:

$$ S = \sum_{k=0}^n {{n}\choose{k}} a^{n-k}(bi)^{k} + \sum_{k=0}^n {{n}\choose{k}} a^{n-k}(-bi)^{k} $$

Now note that whenever k is odd, $(bi)^k = - (-bi)^k $ so that the term from the first sum exactly cancels the term from the second sum.

If n is even, S can then be expressed as

$$ S = 2 \sum_{k=0}^{n/2} {{n}\choose{2k}} a^{n-2k}(bi)^{2k} $$

and if n is odd, then

$$ S = 2 \sum_{k=0}^{(n-1)/2} {{n}\choose{2k}} a^{n-2k}(bi)^{2k} $$

We can simplify these expressions now so that

for even n: $$ S = 2 \sum_{k=0}^{n/2} {{n}\choose{2k}} a^{n-2k}(b)^{2k}(-1)^{k} $$

and for odd n:

$$ S = 2 \sum_{k=0}^{(n-1)/2} {{n}\choose{2k}} a^{n-2k}(b)^{2k}(-1)^k $$

Which are expressions for the sum with no included factors of i.

wanderingmathematician
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