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# Complex powers

A thought popped into my head the other day:

$\text{What would you get when you put a complex number to the power of another complex number?}$

So I decided to try and figure it out algebraically, here are the fruits of my labour.

Let $$z_n$$ be a complex number in the form $$a_n + b_ni$$ where $$a_n$$ and $$b_n$$ are real numbers.

$z_1 = a_1 + b_1i,~ z_2 = a_2 + b_2i$

$\large z_1^{z_2} = (a_1 + b_1i)^{(a_2 + b_2i)}$

$z_n = r_ne^{\theta_ni}$

$\large (r_1e^{\theta_1i})^{(r_2e^{\theta_2i})}$

$\large r_1^{(r_2e^{\theta_2i})} \cdot (e^{\theta_1i})^{(r_2e^{\theta_2i})}$

Let $$cis~ \theta = \cos \theta + i\sin \theta$$

$e^{\theta_ni} = cis~ \theta_n$

$\large r_1^{(r_2cis~ \theta_2)} \cdot (cis~ \theta_1)^{(r_2cis~ \theta_2)}$

$(cis~ \theta_n)^x = cis~ \theta_nx$

$\large r_1^{(r_2cis~ \theta_2)} \cdot cis~ (\theta_1 \cdot (r_2cis~ \theta_2))$

$cis~ (\theta_1r_2cis~ \theta_2)$

$cis~ (\theta_1r_2\cos \theta_2 + \theta_1r_2i\sin \theta_2)$

$cis~ (a + bi) = \frac{cis~ a}{e^b}$

Proof of the above statement here

$\large \frac{cis~ (\theta_1r_2\cos \theta_2)}{e^{(\theta_1r_2\sin \theta_2)}}$

$\large r_1^{(r_2cis~ \theta_2)} \cdot \frac{cis~ (\theta_1r_2\cos \theta_2)}{e^{(\theta_1r_2\sin \theta_2)}}$

$\large r_1^{(r_2\cos \theta_2)} \cdot r_1^{(ir_2sin~ \theta_2)} \cdot \frac{cis~ (\theta_1r_2\cos \theta_2)}{e^{(\theta_1r_2\sin \theta_2)}}$

$\large r_1^{(r_2\cos \theta_2)} \cdot \left(e^{\ln r_1}\right)^{(ir_2sin~ \theta_2)} \cdot \frac{cis~ (\theta_1r_2\cos \theta_2)}{e^{(\theta_1r_2\sin \theta_2)}}$

$\large r_1^{(r_2\cos \theta_2)} \cdot e^{ir_2\sin \theta_2 \ln r_1} \cdot \frac{cis~ (\theta_1r_2\cos \theta_2)}{e^{(\theta_1r_2\sin \theta_2)}}$

$\large r_1^{(r_2\cos \theta_2)} \cdot cis~ (r_2\sin \theta_2 \ln r_1) \cdot \frac{cis~ (\theta_1r_2\cos \theta_2)}{e^{(\theta_1r_2\sin \theta_2)}}$

$e^{xi} \cdot e^{yi} = cis~ x \cdot cis~ y$

$e^{xi} \cdot e^{yi} = e^{xi + yi} = e^{i(x + y)}$

$e^{xi} \cdot e^{yi} = cis~ (x + y)$

$cis~ x \cdot cis~ y = cis~ (x + y)$

$\large r_1^{(r_2\cos \theta_2)} \cdot \frac{cis~ (\theta_1r_2\cos \theta_2 + r_2\sin \theta_2 \ln r_1)}{e^{(\theta_1r_2\sin \theta_2)}}$

$r_1 = \sqrt{a_1^2 + b_1^2},~ r_2 = \sqrt{a_2^2 + b_2^2}$

$\theta_1 = \arctan \frac{b_1}{a_1},~ \theta_2 = \arctan \frac{b_2}{a_2}$

$\large z_1^{z_2} = \sqrt{a_1^2 + b_1^2}^{~\left(\sqrt{a_2^2 + b_2^2}\cos~ \left(\arctan \frac{b_2}{a_2}\right)\right)} \cdot \frac{cis~ \left(\sqrt{a_2^2 + b_2^2}\arctan \frac{b_1}{a_1} \cos \left(\arctan \frac{b_2}{a_2}\right) + \sqrt{a_2^2 + b_2^2}\sin~ \left(\arctan \frac{b_2}{a_2}\right)\ln \sqrt{a_1^2 + b_1^2}\right)}{e^{\left(\sqrt{a_2^2 + b_2^2} \arctan \frac{b_1}{a_1} \sin \left(\arctan \frac{b_2}{a_2}\right)\right)}}$

Well that's a mouthful, so I tried to find a simpler version. What follows is my second attempt.

$\large z_1^{z_2} = \left(r_1e^{\theta_1i}\right)^{(a_2 + b_2i)}$

$\large r_1^{(a_2 + b_2i)} \cdot e^{(a_2\theta_1i - b_2\theta_1)}$

$\large r_1^{a_2} \cdot r_1^{b_2i} \cdot e^{-b_2\theta_1} \cdot e^{a_2\theta_1i}$

$\large r_1^{a_2} \cdot \left(e^{\ln r_1}\right)^{b_2i} \cdot \frac{cis~ (a_2\theta_1)}{e^{b_2\theta_1}}$

$\large r_1^{a_2} \cdot e^{ib_2\ln r_1} \cdot \frac{cis~ (a_2\theta_1)}{e^{b_2\theta_1}}$

$\large r_1^{a_2} \cdot cis~ (b_2\ln r_1) \cdot \frac{cis~ (a_2\theta_1)}{e^{b_2\theta_1}}$

$\large r_1^{a_2} \cdot \frac{cis~ (a_2\theta_1 + b_2\ln r_1)}{e^{b_2\theta_1}}$

$\large \frac{r_1^{a_2}}{e^{b_2\theta_1}} \cdot cis~ (a_2\theta_1 + b_2\ln r_1)$

$\large \frac{\left(e^{\ln r_1}\right)^{a_2}}{e^{b_2\theta_1}} \cdot cis~ (a_2\theta_1 + b_2\ln r_1)$

$\large \frac{e^{a_2\ln r_1}}{e^{b_2\theta_1}} \cdot cis~ (a_2\theta_1 + b_2\ln r_1)$

$\large e^{a_2\ln r_1 - b_2\theta_1} \cdot cis~ (a_2\theta_1 + b_2\ln r_1)$

$\large z_1^{z_2} = e^{a_2\ln \sqrt{a_1^2 + b_1^2} - b_2\arctan \frac{b_1}{a_1}} \cdot cis~ \left(a_2\arctan \frac{b_1}{a_1} + b_2\ln \sqrt{a_1^2 + b_1^2}\right)$

Hope you enjoyed the note.

Note by Jack Rawlin
1 year, 3 months ago