Solving $e^z = 1 $ in complex plane

Question

I am solving the equation $e^z = 1 $ in $\mathbb{C}$. The book says, other than $z = 0$, $z = 2 \pi k i$ for $ k \in \mathbb{Z}$ is also the solution. It explains the solution by saying that $e^z$ is periodic function so that

$1 = e^z = e^{2\pi k i}$

However I want to know how the identity is derived to solve other cases such as $e^z = 2 $.

Answer 1

Note

$$e^z=2=e^{\ln2+2\pi ki}$$

which leads to the solutions

$$z = \ln 2+2\pi ki$$

Answer 2

Let $z=x+iy$ where $x, y$ are real. Then $e^z = 2$ means $$ e^x\cos y + i e^x\sin y = 2, \\ e^x\cos y = 2\quad\text{and}\quad e^x\sin y = 0 $$ Now $e^x \ne 0$ for all $x$, so from $e^x\sin y = 0$ we get $\sin y = 0$, and thus $y = n\pi$ for some $ n \in \mathbb Z$. From this we get $\cos y = \cos(n\pi) = (-1)^n$. But $e^x > 0$ for all $x$, so we must have $n$ even, say $n=2k$. Finally, $2 = e^x\cos y = e^x\cos(2k\pi)= e^x$, so $x=\ln 2$. Conclusion: $z = \ln 2 + i2k\pi$.

Answer 3

You want to use the fact that $e^a = e^b$ if and only if $b-a = 2\pi ki$ for some integer $k$.

So as long as you can find some value of $z$ with $e^z=2$, this lets you find all such values.

For this example, you certainly know that $2 = e^{\ln 2}$ (where "$\ln$" is the ordinary real-valued function of a positive real variable), so $$2 = e^z\tag{given equation to solve}$$ $$\iff e^{\ln 2} = e^z\tag{because $2=e^{\ln 2}$}$$ $$\iff z - \ln 2 = 2\pi ki\tag{rule I cited}$$ $$\iff z = \ln 2 + 2\pi ki\tag{solve for $z$}$$ for some integer $k$.

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Now why is the rule true? $e^a = e^b$ is equivalent to $e^{b-a} = 1$ (multiply both sides by $e^{-b}$).

So the rule is really the same as $e^w= 1$ if and only if $w = 2\pi k i$ for some integer $k$.

To see why this is true, remember that for real $u$ and $v$, we have $e^{u + iv} = e^u(\cos v + i\sin v)$.

This is $1$ precisely when $e^u=1$ (so $u=0$) and $(\cos v,\sin v) = (1,0)$ (so $v = 2\pi k$).

In other words, when $u+iv = 0 + 2\pi k i$, as required.

Answer 4

There is a change in the modulus only. If $z=x+iy$, $e^z=e^xe^{i\theta}=1$ and in the second case $e^z=e^xe^{i\theta}=2.$ You can observe that no change in the arguments. That is why In the first case $e^x=1; x=\ln 1=0, y=2k\pi$ and in the second case $e^x=2; x=\ln 2, y=2k\pi.$ Note that $|e^z|=e^x$