Our job is to find $z$, where $z\in\Bbb{C}$.
What we know: $$|z|=\sqrt{3}$$ $$Re(z)=\frac{\sqrt{2}}{2}$$
My attempt
Since all complex numbers are in the form $z=a+bi$, $a$ being a real part and $b$ being an imaginary part, that is $Re(z)=a \land Im(z)=b$, we know: $$Re(z)=\frac{\sqrt{2}}{2}\iff a=\frac{\sqrt{2}}{2}$$ So, we have $a$. We just need to find $b$.
There is an identity $$|z|=z\cdot\bar{z}$$ This can be rewritten as $$|z|=(a+bi)(a-bi)$$ Substituting our values in this identity we get: $$\sqrt{3}=\left(\frac{\sqrt{2}}{2}+bi\right)\left(\frac{\sqrt{2}}{2}-bi\right)$$ If we multiply all factors on the right-hand side we get: $$\sqrt{3}=1/2+b^2$$ From that we have: $$b^2=\sqrt{3}-1/2\iff b=\pm\sqrt{\sqrt{3}-1/2}$$ Therefore, the solution must be $$z=\frac{\sqrt{2}}{2}\pm\sqrt{\sqrt{3}-1/2}i$$
Unfortunately, this isn't the solution my workbook gives. The solution my workbook gives is $$z=\frac{\sqrt{2}}{2}\pm\frac{i\sqrt{10}}{2}$$ Is my solution wrong or just written differently? If it's wrong, where is my mistake and what is the correct way to solve this problem? If it's just written differently, how can I rewrite my solution to match the one given by my workbook? Also, it's possible that my solution is correct while the one in the workbook is wrong, altough very unlikely, but if that turns out to be the case please let me know.
Your solution is wrong because $z.\overline z=|z|^2$. Correct what you did based on this and you will get the correct solution.