Finding "hidden" solutions to a simple complex system of linear equations

Question

I have been trying to solve the following system of linear equations in the complex plane:

$$\begin{cases} z_1 = -iz_2 \\ z_2 = iz_1 \end{cases} $$ I know the solution, it's $z_1 = 1, \space z_2 = i$, but i can't find a way to prove it, it seems like the solution is "hidden". If someone could provide a proof (and maybe an explanation on what's going on), I would much appreciate it.

Answer 1

It's easiest to see when you bring everything on one side and order by variables:

$z_1+iz_2=0$

$-iz_1+z_2=0$.

Now, multiply both sides of the second equation by $i$. You get:

$z_1+iz_2=0$

$z_1+iz_2=0$.

That's twice the same equation! What this means is that you really only have one distinct equation for two variables, i.e. you are undetermined. Again, what this means is that you can choose any value for $z_1$, which then fixed $z_2=i z_1$.

Answer 2

Hint

$e^{i\frac{\pi}2}=i\Rightarrow\frac 1i=e^{-i\frac{\pi}2}=-i\rightarrow\boxed{\frac 1i=-i}$

Answer 3

Both the equations $z_1 = -iz_2$ and $z_2 = iz_1$ are the same equation and can be converted in to each other.

Let, $$z_1=r_1(\cos\varphi+i\sin\varphi)=r_1e^{i\varphi}$$ $$z_2=r_2(\cos\phi+i\sin\phi)=r_2e^{i\phi}$$

Using $z_1 = -iz_2$ $$r_1e^{i\varphi}=-r_2e^{i\phi}$$

Using Euler's formula $e^{-i\frac{\pi}{2}}=-i$ $$r_1e^{i\varphi}=r_2e^{i\bigl(\phi-\frac{\pi}{2}\bigr)}$$

Also, magnitude of both $z_1$ and $z_2$ is same $\implies$ $r_1=r_2$

$\therefore$ There may be an infinite solution to this problem because of relation $\varphi=\phi-\frac{\pi}{2}$ and this relation still holds true if the magnitude of both $z_1$ and $z_2$ is same $\implies$ $r_1=r_2$.