Solve $z^2+iz=0$

Question

I need to solve $$z^2+iz=0$$.

I need to explicit trigonometric and algebric form.
Since there is a i in the equation I'm not able to start with $z=\rho( \cos\theta +i\sin\theta)$ so I used:

$$z = (a + ib )$$

and I get:

$$(a+ib)^2 + i(a-ib) = 0$$ $$a^2-b^2+2aib+ia+b = 0$$

but now I don't know how to procede and, is this the rigth way?

Answer 1

There's an easier way:

$$z^2 + iz = 0$$

$$z(z+i)=0$$

$$z=0\quad\mathrm{or}\quad z+i=0$$

So the two solutions are $z=0$ and $z=-i$.

Answer 2

$z(z+i) = 0$

Since $C $ is a field, $z = 0$ or $z = -i$

So $z = cos(\pi/2)+sin(\pi/2)i$ or $z = 0$.

When you get $a^2-b^2+2aib+ia+b = 0$ , the real part and imaginary part must be 0. So $a = 2a \implies a = 0$, and $ b = 1 $ or $b = 0$.

Answer 3

$$z^2+iz=0\iff z(z+i)\iff z\in\{0,-i\},$$ the rest follow.

Answer 4

While the factorisations in the other answers are nice to see, don't forget: You are allowed to use the usual solution formula $\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$ for quadratic equations (as long as you're careful with the complex square root), so the answer is: $$ z = \frac{-i \pm \sqrt{i^2 - 4\cdot 1 \cdot 0}}{2} = \frac{-i\pm i}{2} $$ which gives the two solutions $-i$ or $0$.