$z$ is a complex number. I have to write its solution(s) in the form $z= a +bi$ given that :
$$ z^2 + 1 - i = 0 $$
I put $1 -i$ on the other side and took the root. That would give $z= \sqrt{-1+i}.$ I tried to manipulate $ -1+ i,$ obtaining $ z=i \sqrt{1+i^3} $ but that doesn't help me further.
Thanks for your help.
Once you have,
$$z^2 = i - 1$$
Let $z = a + ib$,
$$(a+ib)^2 = i - 1$$
Expand the brackets, and see what system of equations you can form to find $a,b$. An alternative would be to write $i-1$ in polar form, but I'm not sure what level you're working at.
Hint:
Let $z=a+ib$ $$z^2=a^2-b^2+2iab$$
Compare to get $$2ab=1 \text{ and }a^2-b^2=-1$$
Hope you can continue after this?
First there is no single square root of a complex number, so the notation $\sqrt z$ is meaningless, unless you define a branch cut on $\mathbf C$. For positive real numbers, the situation is different, as one decides, conventionally, that the symbol $\sqrt x$ denotes the non-negative square root of $x$.
This being said, observe $\; -1+i=\sqrt 2\,\mathrm e^{\tfrac{3i\pi}4}$, so its square roots are $$\pm\sqrt[4] {\mathstrut2}\,\mathrm e^{\tfrac{3i\pi}8}=\pm\sqrt[4] {\mathstrut2}\Bigl(\sin\dfrac\pi8+i\cos\dfrac\pi8\Bigr).$$
Another approach not covered here yet is to use the polar form of the complex number. Note that $$ i - 1 = \sqrt2 \left[ \cos(3\pi/4) + i \sin(3\pi/4)\right] = \sqrt{2}e^{3\pi i/4} $$ Now you can take square roots (be careful with multiple branches) and convert back to rectangular form using Euler's formula I just used above to convert from rectangular to polar...
$$\sqrt{-1+i}=\sqrt[4]{2}\sqrt{\cos135^{\circ}+i\sin135^{\circ}}=$$ $$=\{\sqrt[4]2\left(\cos67.5^{\circ}+i\sin67.5^{\circ}\right),\sqrt[4]2\left(\cos247.5^{\circ}+i\sin247.5^{\circ}\right)\}$$ $$=\{\sqrt[4]2\left(\sin22.5^{\circ}+i\cos22.5^{\circ}\right),\sqrt[4]2\left(-\sin22.5^{\circ}-i\cos22.5^{\circ}\right)\}=$$
$$=\left\{\sqrt[4]2\left(\sqrt{\frac{1-\frac{1}{\sqrt2}}{2}}+\sqrt{\frac{1+\frac{1}{\sqrt2}}{2}}i\right),\sqrt[4]2\left(-\sqrt{\frac{1-\frac{1}{\sqrt2}}{2}}-\sqrt{\frac{1+\frac{1}{\sqrt2}}{2}}i\right)\right\}=$$ $$=\left\{\sqrt{\frac{\sqrt2-1}{2}}+\sqrt{\frac{\sqrt2+1}{2}}i,-\sqrt{\frac{\sqrt2-1}{2}}-\sqrt{\frac{\sqrt2+1}{2}}i\right\}.$$
