0
$\begingroup$

Introduction

For a few months I have been using calculations of imaginative powers and I have come across the equation that most of you are familiar with by now being:

$$x^{\left(y\cdot i\right)}=\cos \left(y\cdot \ln \left(x\right)\right)+i\cdot \sin \left(y\cdot \ln \left(x\right)\right)$$

$x$ and $y$ are both considered as variables in this instance.

Question

How is this equation come about, I have a feeling it is something to do with $e$ being the derivative of itself, however, I haven't done any calculus yet at my school (since in the United Kingdom, Calculus isn't until A level).

I was wondering if anyone would be able to give a helpful explanation of why this is true, not that I'm doubting it, just looking for some more knowledge.

I have actually created a desmos graph for any of you who are interested, it uses real numbers and maps them onto an imaginary plane. Here is the link:

$x^{\left(y\cdot i\right)}$ graph

Hope that you like the graph and hopefully you can help me out with this question.

$\endgroup$
  • $\begingroup$ Wait, so have you learned the taylor expansion series yet? $\endgroup$ – Crescendo Aug 20 '17 at 15:14
3
$\begingroup$

That is because of how the exponential function behaves on complex plane: $$ e^{a+ib} = e^a (\cos(b) + i \sin(b)) $$ This identity can be proved in various ways.

$\endgroup$
1
$\begingroup$

The usual definition of exponentiation for real or complex numbers is that $x^y=e^{y \ln(x)}$. In the real numbers this is fine. In the complex numbers the natural log is multivalued as you can add $2\pi i$ to one value of the logarithm and get another. Once you choose the branch of the log, we have $x^{yi}=e^{yi\ln (x)}$ We can then plug that into Euler's formula $e^{iz}=\cos(z)+i \sin(z)$to get $$x^{yi}=\cos(y \ln(x))+i\sin(y\ln(x))$$

$\endgroup$
0
$\begingroup$

We can use the expansion$$a^x=1+x\log a+\frac {x^2\log^2a}{2!}+\frac {x^3\log^3a}{3!}+\ldots$$to expand $x^{yi}$ and show that the expansion is equivalent to $\cos\left(y\log x\right)+i\sin\left(y\log x\right)$. The formula above can be proven using Newton's generalized binomial theorem, and a bit of algebraic manipulation. Replacing $x$ with $yi$ and $a$ with $x$, we get$$\begin{align*}x^{yi} & =1+yi\log x-\frac {y^2\log^2x}{2!}-\frac {y^3i\log^3x}{3!}+\frac {y^4\log^4x}{4!}+\ldots\\ & =\left(1-\frac {y^2\log^2x}{2!}+\frac {y^4\log^4x}{4!}-\ldots\right)+i\left(y\log x-\frac {y^3\log^3x}{3!}+\frac {y^5\log^5x}{5!}-\ldots\right)\\ & =\cos\left(y\log x\right)+i\sin\left(y\log x\right)\end{align*}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.