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I'm trying to develop a simple program to evaluate integral powers of a complex number \$z\$ , that is, \$z^n\$, where \$z\$ is in the algebrical form \$a+i b\$ and \$n \in \mathbb{Z} ^{*} _{+}\$.

How I was trying to proceed:

First, I defined a product function, and next I intended to define a exponentiation function inside of which I iterated the product function (the code can bee seen below). Is it correct? Is there a more efficient alternative for doing that?

    include <stdio.h>
    include <math.h>

    struct complex {
        float real,imag;
    };

    struct complex product (struct complex x, struct complex y){
    /* using the distributive property, (a+ib)(c+id) = (ac-bd)+i(ad+bc)*/
        struct complex z;
        z.real = (x.real * y.real)- (x.imag * y.imag);
        z.imag = (x.real * y.imag) + (x.imag * y.real);
    return z;
    }      

    struct complex exponentiation (struct complex z, int n){
        int i;
        struct complex w;
        w = z;
        for (i=1, i<n, i++)
            w = product (w,z);
    return w;
    }
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2 Answers 2

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C has a built-in complex type and a cpow() function since C99. You seem to be .

Only in rare circumstances should you use float. In most cases, you should use double instead.

Exponentiation of complex numbers is better done using the polar representation:

$$z = r e^{i\theta}$$

where

$$\begin{align} z &= a + i b \\ r &= \sqrt{a^2 + b^2} \\ \theta &= \arctan{\frac{b}{a}} \end{align}$$

then

$$z^n = (r e^{i\theta})^n = r\,^n e^{i\ n \theta}$$

For non-trivial values of \$n\$, you would be better off converting to a polar representation, performing the exponentiation, and converting back to the Cartesian representation.

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  • Your algorithm is linear with respect to the exponent. It could (and should) be done in a logarithmic time.

  • complex exponentiation (struct complex z, int n) as coded returns wrong answer for non-positive n.

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  • \$\begingroup\$ To be fair, it is declared in the question that n ∈ ℤ∗+. Perhaps the problem is that it doesn't validate that n > 0. \$\endgroup\$ Commented Mar 3, 2015 at 8:36

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