Find all complex numbers $z$ such that $e^z=1+2i$

Question

I'm not sure how to do this. I would really love it if someone would talk me through it. I know we must somehow solve $$ z=\log1+2i $$ Another idea I had is that we know that $e^{2\pi i}=1$ and $e^{\pi i/2}=i$, so $$ 1+2i=e^{2\pi i}+2e^{\pi i/2} $$ and maybe this would help but I don't yet see how.

Answer 1

We know by Euler's Formula that $$re^{i\theta} = r\cos\theta +ir\sin\theta$$$r$ will be equal to the magnitude of the complex number, i.e., $$\sqrt{\textrm{Re}(z)^2 + \textrm{Im}(z)^2} = \sqrt 5$$So, we look for all $\theta$ such that $\cos\theta = \frac1{\sqrt 5}$ and $\sin\theta = \frac2{\sqrt 5}$. Can you solve this from here?

Answer 2

Use that $$e^z=e^x\cos(y)+ie^x\sin(y)$$ if $$z=x+iy$$ and you will get

$$e^x\cos(y)-1+i(e^x\sin(y)-2)=0$$

Answer 3

First of all, note that\begin{align}1+2i&=\sqrt5\left(\frac1{\sqrt5}+\frac2{\sqrt5}i\right)\\&=\sqrt5e^{\arccos\left(\frac1{\sqrt5}\right)i}\\&=e^{\log\sqrt5+\arccos\left(\frac1{\sqrt5}\right)i}.\end{align}So, now you have one solution of the equation $e^z=1+2i$:$$z=\log\sqrt5+\arccos\left(\frac1{\sqrt5}\right)i.$$And I suppose that you know that $e^z=1$ if and only if $z=2\pi ni$, for some integer $n$. Now, put it all together…

Answer 4

We note that

$1 + 2i = \sqrt 5 \left (\dfrac{1}{\sqrt 5} + i\dfrac{2}{\sqrt 5} \right); \tag 1$

set

$z = x + iy; \tag 2$

then

$e^z = e^{x + iy} = e^x e^{iy} = e^x(\cos y + i \sin y) = \sqrt 5 \left (\dfrac{1}{\sqrt 5} + i\dfrac{2}{\sqrt 5} \right); \tag 4$

thus the modulus of $e^z$ is

$\vert e^z \vert = \vert e^x(\cos y + i \sin y) \vert = \vert e^x \vert \vert \cos y + i \sin y \vert = e^x = \sqrt 5; \tag 5$

the only real $x$ satisfying this equation is

$x = \ln \sqrt 5 = \dfrac{1}{2} \ln 5; \tag 6$

returning to (4) and using (5):

$e^{iy} = \cos y + i \sin y = \dfrac{1}{\sqrt 5} + i \dfrac{2}{\sqrt 5}; \tag 7$

it follows from (7) that there is a unique

$y_0 \in \left [0, \dfrac{\pi}{2} \right ] \tag 8$

with

$\tan y_0 = \dfrac{\sin y_0}{\cos y_0} = 2 \tag 9$

and

$e^{iy_0} = \cos y_0 + i\sin y_0 = \dfrac{1}{\sqrt 5} + i \dfrac{2}{\sqrt 5}; \tag{10}$

setting

$z = x + iy \tag{11}$

with $x$ and $y$ as in (5)-(6) and (10), respectively, we write

$z = \ln \sqrt 5 + iy_0; \tag{12}$

we check (12):

$e^z = e^{\ln \sqrt 5 + iy_0} = e^{\ln \sqrt 5}e^{iy_0} = \sqrt 5 ( \cos y_0 + i \sin y_0) = \sqrt 5 \left ( \dfrac{1}{\sqrt 5} + i \dfrac{2}{\sqrt 5} \right ) = 1 + 2i; \tag{13}$

having found $x \in \Bbb R$ and $y_0 \in [0, 2\pi]$ satisfying (13), we may search for other solutions. If $w = \sigma + i\omega \in \Bbb C$ is such that

$e^{\sigma + i\omega} = e^w = e^z = e^{x + iy_0} = 1 + 2i, \tag{14}$

then it follows exactly as above that

$e^\sigma = \sqrt 5 = e^x, \; \sigma = \ln \sqrt 5 = x; \tag{15}$

then

$e^{i\omega} = e^{iy_0} = \dfrac{1}{\sqrt 5} + i \dfrac{2}{\sqrt 5} \tag{16}$

or

$e^{i(\omega - y_0)} = 1; \tag{17}$

it follows that

$\omega - y_0 = 2n\pi, \; n \in \Bbb Z, \tag{18}$

whence

$\omega = y_0 + 2n \pi, \; n \in \Bbb Z; \tag{19}$

it is easy to see that every $\omega$ given by (19) satisfies

$e^{i\omega} = e^{i(y_0 + 2n\pi)} = e^{iy_0} e^{2n \pi i} = e^{iy_0}; \tag{20}$

we have seen that (19) is both necessary and sufficient for

$e^z = e^{x + iy} = e^{x + i\omega} = 1 + 2i; \tag{21}$

thus every solution is of the form

$z = \ln \sqrt 5 + i(y_0 + 2n\pi). \tag{22}$