Solve the equation $z^{4} + (1-i)z^{2} - i = 0$ for $z\in\mathbb{C}$

Question

The proposed equation is given by:

$$z^{4} + (1-i)z^{2} - i = 0$$

MY ATTEMPT

I tried to start solving it like this:

\begin{align*} (a+bi)^4 + (i - 1)(a + bi)^2 -i = 0 & \Longleftrightarrow a^4 + 4a^3bi - 6a^2b^2 - 4ab^3i + b^4 = 0\\\\ & \Longleftrightarrow a^4 + b^4 = 2ab(3ab + 2b^2i - 2a^2i) \end{align*}

But I couldn't do it further.

Answer 1

HINT

Here it is an alternative way to solve it for the sake of curiosity.

\begin{align*} z^{4} + (1-i)z^{2} - i = 0 & \Longleftrightarrow (z^{4} + z^{2}) - i(z^{2} + 1) = 0\\\\ & \Longleftrightarrow z^{2}(z^{2} + 1) - i(z^{2} + 1) = 0\\\\ & \Longleftrightarrow (z^{2} - i)(z^{2} + 1) = 0 \end{align*}

Can you take it from here?