Continuing from your method (making sure to use a new letter), we have $$ \frac{a^2-b^2-2abi}{a+bi} = 0 + ic$$
Simplifying gives us $a^2-b^2 - 2abi = -bc + aci$. It looks very unwieldy, but let's use our usual techniques (comparing real and imaginary parts). Our final aim is to find $w$ in terms of $a$ so let's find $b$ and $c$ in terms of $a$. Give it a try first, and you can check against my working below:

Comparing the imaginary parts, $-2ab = ac$. Since $a \neq 0, c = -2b$.

Comparing the real parts, $a^2-b^2 = -bc$. Substituting $c=-2b$ gives us $a^2 = 3b^2$ so $b \displaystyle = \pm \frac{a}{\sqrt{3}}$

Hence $\displaystyle w = a + i \frac{a}{\sqrt{3}}$ or $\displaystyle w = a - i \frac{a}{\sqrt{3}}$

Alternative method

We have already calculated that $\displaystyle \frac{(w^*)^2}{w} = \frac{a^2-b^2-2abi}{a+bi}$. Notice there is a division in our working, so we can further simplify ("rationalize" using conjugates)! Give it a try.