Find all numbers z (if any) such that $e^{6z}=2i$.

Question

I'm not really sure how I should go about solving this. This is a question from the section before logs, so we're supposed to solve it with the knowledge we have of exponential, trig, and hyperbolic functions and not logs. I know that $e^z=1$ iff $z=2k\pi i$ for some integer z. I also know that $e^{z_1}=e^{z_2}$ holds iff $z_1=z_2 + 2k\pi i$ for some integer k.

I've tried something like letter $2i=e^x$ and then $e^{6z}=e^x => 6z=x$, where x is just some unknown at this point, but I'm feeling like I'm missing some key piece of information.

Answer 1

We know that $e^{i\theta}=\cos\theta+i\sin\theta$ so that $|e^{i\theta}|=1.$ If we write $z=x+iy$, with $x,y\in\mathbb{R}$, then$$|e^{x+iy}|=|e^x||e^{iy}|=e^x$$ Therefore, if we want to solve $e^{6z}=2i$ we must have $\Re(6z)=\ln2$. If $\theta=\Im(6z)$ then $e^{i\theta}=i$. One possible value of $\theta$ is $\frac\pi2$ and we get another by adding any multiple of $2\pi i$. That is, $$6z=\ln2+\left(2k+\frac12\right)\pi i,\ k=0,\pm1,\pm2,\dots$$or $$z=\frac{\ln2}{6}+\frac{4k+1}{12}\pi i,\ k=0,\pm1,\pm2,\dots$$

Answer 2

So if $z=x+iy$

$$e^{6x} \cdot e^{6iy}=e^{\ln{2}} \cdot e^{\frac{iπ}{2}}$$

$x=\dfrac{\ln{2}}{6}$

$iy=\dfrac{iπ}{12}+\dfrac{πn}{3} \implies n \in \mathbb{N}$

$z=\dfrac{\ln{2}}{6}+(\dfrac{π}{12}+\dfrac{πn}{3})i$

Does Euler's equation count?

Edit: I forgot to add in the periodicity of the complex exponential. But it's not $2\pi$ directly as you might expect. It's periodic for $6iy$ so the number itself shares in only a sixth of that.

Answer 3

I'm pretty sure this question cannot be answered without referring to the function $\ln x$ for real $x$; see below. On the other land, the intricacies of the complex logarithm $\ln z$ may be avoided via the Euler formula

$e^{i\theta} = \cos \theta + i \sin \theta, \; \theta \in \Bbb R, \tag 0$

which implies

$e^{i\theta} = 1 \Longleftrightarrow \theta = 2\pi n, \; n \in \Bbb Z; \tag{0.5}$

see (11)-(12) below.

Now with

$z = x + iy, \; x, y \in \Bbb R;, \tag 1$

we have

$6z = 6x + 6iy = 6x + 6yi, \tag 2$

whence

$e^{6z} = e^{6x + 6yi} = e^{6x}e^{6yi}; \tag 3$

with $\theta = \pi / 2$, (0) implies

$i = e^{\pi i/2}, \tag 4$

and so we proceed as follows:

$e^{6z} = 2i \Longrightarrow e^{6x + 6yi} = 2i \Longrightarrow e^{6x}e^{6yi} = 2e^{ \pi i/2}; \tag 5$

since

$0 < e^{6x} \in \Bbb R, \tag{5.5}$

$e^{6x} = \vert e^{6x} \vert = \vert e^{6x} \vert \vert e^{6yi} \vert = \vert e^{6x}e^{6yi} \vert $ $= \vert 2e^{\pi i/2} \vert = 2\vert e^{\pi i / 2} \vert = 2; \tag 6$

with $x$ real this yields

$6x = \ln 2, \tag 8$

or

$x = \dfrac{\ln 2}{6}. \tag 9$

By virtue of (5) and (6),

$e^{6yi} = e^{\pi i/ 2}, \tag{10}$

that is,

$e^{(6y - \pi/2)i} = e^{6yi - \pi i/2} = 1, \tag{11}$

$\left (6y - \dfrac{\pi }{2}\right)i = 2\pi i n, \; n \in \Bbb Z, \tag{12}$

$6y = \dfrac{\pi}{2} + 2\pi n, \; n \in \Bbb Z, \tag{13}$

$y = \dfrac{\pi}{12} + \dfrac{n \pi}{3}, \; n \in \Bbb Z; \tag{14}$

finally,

$z = x + iy = \dfrac{\ln 2}{6} + \left ( \dfrac{\pi}{12} + \dfrac{n \pi}{3} \right ), \; n \in \Bbb Z. \tag{15}$

Back-tracking through (15) to (4) provides the steps necessary to check this result.