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I was solving practice problems for my upcoming midterm and however I got stuck with this question type.

It is asking me to find all roots and then sketch it.

$(1+isqrt3)^1/2$

How do we proceed?

Hint:Substitute $$sqrt1+sqrt3i=a+bi$$ then you have to solve $$a^2-b^2=1$$ and $$sqrt3=2ab$$

Let the roots be of the form $u+iv$. You need to solve

$$(u+iv)^2=1+isqrt3.$$

As $(u+iv)^2=u^2-v^2+2iuv$, by identification

$$begincasesu^2-v^2=1,\2uv=sqrt3endcases.$$

Now multiplying the first by $u^2$,

$$u^4-u^2v^2=u^4-frac34=u^2$$ is a biquadratic equation. We obtain

$$u^2=frac1pmsqrt1+32=frac32,-frac12$$

and the only real roots are

$$u=pmfracsqrt3sqrt2,$$ corresponding to $$v=pmfrac1sqrt2.$$

Final check:

$$left(fracsqrt3sqrt2+frac isqrt2right)^2=frac32-frac12+2ifracsqrt32$$

$1+isqrt3=2e^ifracpi3=(sqrt2e^ifracpi6)^2$ thus the roots are $pm sqrt2e^ifracpi6=pm(fracsqrt62+ifracsqrt22)$

General way to find the roots of $z=a+ib$

Let $delta=x+iy$ a root of $z$. Then $delta^2=x^2-y^2+2ixy=a+ib$ and thus $x^2-y^2=a$ and $2xy=b$. Using modulus we obtain : $|z|=|delta|^2Longrightarrow x^2+y^2=sqrta^2+b^2$.

This provides the following system:
$$begincasesx^2-y^2=a ,\x^2+y^2=sqrta^2+b^2,\2xy=b endcases.$$

The first two equations gives $$x^2=fraca+sqrta^2+b^22$$ $$ y^2=fraca-sqrta^2+b^22.$$

$$x=pmsqrtfraca+sqrta^2+b^22, y=pmsqrtfraca-sqrta^2+b^22.$$

The last equation can be used to find the sign of $x$ and $y$. In your example $2xy=sqrt3>0$ which means that $sign(x)=sign(y)$ and thus we have to choose $x>0$ and $y>0$, or $x<0$ and $y<0$.