# How to raise a complex number to the power of another complex number?

How do I calculate the outcome of taking one complex number to the power of another, ie $\displaystyle {(a + bi)}^{(c + di)}$?

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This question is related to this math.stackexchange.com/questions/9770/… – Américo Tavares Nov 11 '10 at 0:32
– Carl Mummert Nov 11 '10 at 3:08

First you need to realize that this is a multi-valued function.

Let us choose the principal logarithm for convenience. So our argument $\theta$ will lie between $(-\pi,\pi]$.

We then write $a+ib = r e^{i \theta} = e^{(ln(r) + i \theta)}$.

Now, we get $(a+ib)^{c+id} = e^{(ln(r) + i \theta)(c+id)}$. Do the necessary algebraic manipulations in the exponent to get $e^{(cln(r) - d \theta) + i(d ln(r) + c \theta)}$. You might also want to take a look at the previous question asked on a similar topic.

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Well, assuming principal values of the complex logarithm (otherwise much craziness ensues):

$$(a+bi)^{c+di}=\exp((c+di)\ln(a+bi))$$

$$=\exp((c+di)(\ln|a+bi|+i\arg(a+bi)))$$

$$=\exp((c\ln|a+bi|-d\arg(a+bi))+i(c\arg(a+bi)+d\ln|a+bi|))$$

$$=\exp((c\ln|a+bi|-d\arg(a+bi)))\exp(i(c\arg(a+bi)+d\ln|a+bi|))$$

and I'll let you finish it at this point, using the fact that $\exp(ix)=\cos\;x+i\sin\;x$

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I transcribe part of my answer to this question.

The complex exponential $e^z$ for complex $z=x+iy$ preserves the law of exponents of the real exponential and satisfies $e^0=1$.

By definition

$$e^z=e^{x+iy}=e^xe^{iy}=e^x(\cos y+\sin y)$$

which agrees with the real exponential function when $y=0$. The principal logarithm of $z$ is the complex number

$$w=\text{Log }z=\log |z|+i\arg z$$

so that $e^w=z$, where $\arg z$ (the principal argument of $z$) is the real number in $-\pi\lt \arg z\le \pi$, with $x=|z|\cos (\arg z)$ and $y=|z|\sin (\arg z)$.

The complex power is

$$z^w=e^{w\text{ Log} z}.$$

In your case you have: $z=a+bi,w=c+di$

$$\begin{eqnarray*} \left( a+bi\right) ^{c+di} &=&e^{(c+di)\text{Log }(a+bi)} \\ &=&e^{(c+di)\left( \ln |a+bi|+i\arg (a+bi)\right) } \\ &=&e^{c\ln \left\vert a+ib\right\vert -d\arg \left( a+ib\right) +i\left( c\arg \left( a+ib\right) +d\ln \left\vert a+ib\right\vert \right) } \\ &=&e^{c\ln \left\vert a+ib\right\vert -d\arg(a+bi)}\times \\ &&\times \left( \cos \left( c\arg \left( a+ib\right) +d\ln \left\vert a+ib\right\vert \right) +i\sin \left( c\arg \left( a+ib\right) +d\ln \left\vert a+ib\right\vert \right) \right). \end{eqnarray*}$$

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when you expand $\text{Log}(a+bi)$, shouldn't it be $\log |a+bi|+i \arctan\frac{b}{a}$? This came up in this question: math.stackexchange.com/questions/56384/… – Ross Millikan Aug 8 '11 at 21:00
@Ross Millikan: I corrected $\text{Log }(a+bi)$. I hope now everything is corrected. Many thanks! – Américo Tavares Aug 8 '11 at 21:51
@Ross Millikan: I dedided to correct this answer once more, because $\arg(a+bi)$ may be different to $\arctan b/a$. After this change the final formula is equal to J. M.'s. – Américo Tavares Aug 9 '11 at 10:22
... different from ... – Américo Tavares Aug 9 '11 at 10:51
"different to" makes you sound British-and you are closer to there than to America. – Ross Millikan Aug 9 '11 at 12:47