How to solve $e^{ix}=i$?

Question

This is a question related to another posted question:

The answer to the following question "Find all solutions to: $e^{ix}=i$" is as follows:

"Euler's formula: $e^{ix}=\cos(x)+i\sin(x)$,

so: $ \cos x+i\sin x=0+1⋅i$

compare real and imaginary parts $\sin(x)=1$ and $\cos(x)=0$

$x=\frac{(4n+1)π}2$, $n∈$ (W stands for set of whole number W={0,1,2,3,.......,n})."

My question: Where does $x=\frac{(4n+1)π}2$, $n∈$ come from?

My steps:

1. $\cos(x) + i\sin(x) = 0 + i(1)$

2. $\cos(x) = i(1 - \sin(x))$

3. ...

4. how does $x=\frac{(4n+1)π}2$ follow?

Answer 1

You said it, you compare real and imaginary parts. $\cos(x)=0$ and $\sin(x)=1$. When is this true?

Answer 2

I always find it easier to use a fixed method, and I thought you might find this explanation easier, so I'm posting it.

Start by putting everything into exponential form. Now $i = e^{\frac{i\pi}{2}}$. You can derive this from $e^{i\pi} = -1$ and taking square roots on both sides.

Now note that for any $\theta$, $e^{i\theta} = e^{i(\theta + 2k\pi)}, k \in \mathbb{Z}$, and this is because $e^{2k\pi i} = 1$. Essentially, this can be viewed as the periodicity of the exponential form. To compute general solutions or roots, you would be well-advised to include this term so that you don't miss any solutions.

Hence you can now write $i = e^{i(\frac{\pi}{2} + 2k\pi)} = e^{i\pi\frac{4k+1}{2}}$

Note that the final step is just an algebraic rearrangement of the exponent.

You can now immediately solve the equation by taking logs of both sides, i.e.

$e^{ix} = e^{i\pi\frac{4k+1}{2}}\\ \implies x = \pi\frac{4k+1}{2}$ which is essentially the required form.

Answer 3

$\sin$ and $\cos$ functions are $2\pi$-periodic which is: $\sin(x+2n\pi)=\sin x$, $\cos(x+2n\pi)=\cos(x)$. So when you find that $x=\pi/2$ is a solution, then also $x_n = \pi/2 + 2n\pi$ is a solution for every $n\in \mathbb Z$ (where $\mathbb Z$ are whole numbers: $0, 1, -1, 2, -2,\dots$)

Notice that $$ \frac \pi 2 + 2n\pi = \frac{4n+1}{2}\pi. $$

Answer 4

Hint:

• $i$ is a point on the unit circle.

• $e^{ix}$ is also a point on the unit circle, lying x radians away from $(1,0)$, in trigonometric or counterclockwise direction.

So, to answer a question with a question, What is the position of $i$ on the unit circle, and How many radians away from $(1,0)$ does it lie, in trigonometric or counterclockwise direction ?