Complex power of a complex number: Find $x$ and $y$ in $x + yi = (a + bi)^{c+di}$

Question

$$ x + yi = (a + bi)^{c+di} $$

Find $x$ and $y$ in terms of $a$, $b$, $c$ and $d$.
Where, $i$ is defined as $\sqrt{-1}$ and $a$, $b$, $c$, $d$ are real numbers.

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I defined two new real number variables

$$ r = \sqrt{a^2 + b^2} \\ \theta = arg(a + bi). $$

The expression becomes

$$ x + yi = \left(re^{\theta i}\right)^{c+di} = r^{c+di}e^{-\theta d + \theta c i} = \left[r^{c}e^{-\theta d}\right]r^{di}e^{\theta c i} = \left[r^{c}e^{-\theta d}\right]r^{di}(cos(\theta c) + isin(\theta c)) \\ = r^{di}\left[r^{c}e^{-\theta d}cos(\theta c) + ir^{c}e^{-\theta d}sin(\theta c)\right]. $$

I'm not able to go on any further because of the term $r^{di}$. I don't know how to open it.

Answer 1

Step by step:

$$ \begin{array}{rcl} && x + jy = (a + jb)^{c+jd} \\ &=& (re^{j\theta})^{c+jd} \\ &=& r^{c+jd} (e^{j\theta})^{c+jd} \\ &=& r^cr^{jd} e^{j\theta c}e^{-\theta d} \\ &=& r^c e^{\displaystyle\ln r^{jd}} e^{j\theta c}e^{-\theta d} \\ &=& r^c e^{-\theta d} e^{\displaystyle jd \ln r} e^{\displaystyle j\theta c} \\ &=& r^c e^{-\theta d} e^{\displaystyle j(d \ln r + \theta c)} \\ &=& r^c e^{-\theta d} [\cos(d \ln r + \theta c) + j\sin(d \ln r + \theta c)] \\ &=& [r^c e^{-\theta d} \cos(d \ln r + \theta c)] + j[r^c e^{-\theta d}\sin(d \ln r + \theta c)] \\ &=& \left[(a^2+b^2)^{c/2} e^{-\theta d} \cos\left(d \ln(\sqrt{a^2+b^2}) + \theta c\right)\right] + \\ && \left[(a^2+b^2)^{c/2} e^{-\theta d}\sin\left(d \ln((a^2+b^2)^{1/2}) + \theta c\right)\right]j \\ &=& \left[(a^2+b^2)^{c/2} e^{-\theta d} \cos\left(\frac{d}{2} \ln(a^2+b^2) + \theta c\right)\right] + \\ && \left[(a^2+b^2)^{c/2} e^{-\theta d}\sin\left(\frac{d}{2} \ln(a^2+b^2) + \theta c\right)\right]j \\ &=& \begin{cases} \left[(a^2+b^2)^{c/2} e^{-d\arctan(\frac{b}{a})} \cos\left(\frac{d}{2} \ln(a^2+b^2) + c\,\arctan(\frac{b}{a})\right)\right] + &\\ \left[(a^2+b^2)^{c/2} e^{-d\arctan(\frac{b}{a})} \sin\left(\frac{d}{2} \ln(a^2+b^2) + c\,\arctan(\frac{b}{a})\right)\right]j & \text{if} \; a\!>\!0 \\ \left[(a^2+b^2)^{c/2} e^{-d\arctan(\frac{b}{a}) - \pi d} \cos\left(\frac{d}{2} \ln(a^2+b^2) + c\,\arctan(\frac{b}{a}) + \pi c\right)\right] + & \\ \left[(a^2+b^2)^{c/2} e^{-d\arctan(\frac{b}{a}) - \pi d} \sin\left(\frac{d}{2} \ln(a^2+b^2) + c\,\arctan(\frac{b}{a}) + \pi c\right)\right]j & \text{if} \; a\!<\!0 \; \text{and} \; b \!\ge\! 0 \\ \left[(a^2+b^2)^{c/2} e^{-d\arctan(\frac{b}{a}) + \pi d} \cos\left(\frac{d}{2} \ln(a^2+b^2) + c\,\arctan(\frac{b}{a}) - \pi c\right)\right] + & \\ \left[(a^2+b^2)^{c/2} e^{-d\arctan(\frac{b}{a}) + \pi d} \sin\left(\frac{d}{2} \ln(a^2+b^2) + c\,\arctan(\frac{b}{a}) - \pi c\right)\right]j & \text{if} \; a\!<\!0 \; \text{and} \; b\!<\!0 \\ \left[(a^2+b^2)^{c/2} e^{-\frac{\pi d}{2}} \cos\left(\frac{d}{2} \ln(a^2+b^2) + \frac{\pi c}{2}\right)\right] + & \\ \left[(a^2+b^2)^{c/2} e^{-\frac{\pi d}{2}} \sin\left(\frac{d}{2} \ln(a^2+b^2) + \frac{\pi c}{2}\right)\right]j & \text{if} \; a\!=\!0 \; \text{and} \; b\!>\!0 \\ \left[(a^2+b^2)^{c/2} e^{+\frac{\pi d}{2}} \cos\left(\frac{d}{2} \ln(a^2+b^2) - \frac{\pi c}{2}\right)\right] + & \\ \left[(a^2+b^2)^{c/2} e^{+\frac{\pi d}{2}} \sin\left(\frac{d}{2} \ln(a^2+b^2) - \frac{\pi c}{2}\right)\right]j & \text{if} \; a\!=\!0 \; \text{and} \; b\!<\!0 \\ 0 & \text{if} \; a\!=\!0 \; \text{and} \; b\!=\!0 \\ \end{cases} \end{array} $$

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Implementation in computer:

ComplexNumber ComplexNumber::TakePower(const ComplexNumber & Base, const ComplexNumber & Exponent)
{
    // x + jy = (a + jb)^(c + jd)
    double x, y;
    double a = Base.GetRealPart();
    double b = Base.GetImaginaryPart();
    if (IsZero(a) & IsZero(b))
    {
        x = 0.0;
        y = 0.0;
    }
    else
    {
        double c = Exponent.GetRealPart();
        double d = Exponent.GetImaginaryPart();
        double r2 = a * a + b * b;
        double t = Base.GetAngle();
        x = pow(r2, c/2.0) * exp(-t*d) * cos((d/2.0) * log(r2) + t*c);
        y = pow(r2, c/2.0) * exp(-t*d) * sin((d/2.0) * log(r2) + t*c);  
    }
    return ComplexNumber(x, y);
}

Testing for $e^{j\pi}\!=\!-1$:

double e = 2.71828182845904523536028747135266249775724709369995;
double pi = 3.1415926535897932384626433832795;
ComplexNumber cne(e, 0);
ComplexNumber cni(0, pi);
ComplexNumber cnei = cne.GetPower(cni);
std::cout << "e^(j*pi) = " << cnei.ToString() << std::endl;

Program output:

e^(j*pi) = -1+j1.22465e-016

My implementation calculates $e^{j\pi}$ as $-1\!+\!j1.22465\!\times\!10^{-16}$. I don't know the reason for the error in the imaginary part, but the result is still very close enough to the actual one.

Answer 2

One method is to use Euler's equation. $$ e^{(i\theta)} =cos(\theta)+i sin(\theta)$$ Extending to a general base is messy, but works.

Answer 3

For $\,r,d\in\Bbb R\,$ :

$$r^{di}=e^{di\,\text{Log}(r)}=e^{ri\left(\log|r|+i\arg(r)\right)}$$

Choosing the usual branch for the complex logarithm Log$(z)$ where positive reals have argument equal to zero gives you a result pretty similar to the one would expect from real analysis...

Answer 4

You were pretty close. You can continue like this:

$$(a+ib)^{c+di} = \ldots ... = r^ce^{-d\theta}e^{i(c\theta+d\ln r)}=\\ = r^ce^{-d\theta}\cos(c\theta+d\ln r) + ir^ce^{-d\theta}\sin(c\theta+d\ln r)$$