= Solution
The point $1+i$ has <modulus> $\sqrt2$ and <argument> $\pi/4$ modulo $2\pi$. Taking complex square roots halves the argument and takes the square root of the modulus:
$$
\boxed{|z|=2^{1/4},\qquad \arg z=\frac\pi8+k\pi\pmod{2\pi},\quad k=0,1.}
$$
The two roots are opposite points in the <complex plane>, explicitly $\pm[\sqrt{(\sqrt2+1)/2}+i\sqrt{(\sqrt2-1)/2}]$.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2005/paper-2-complex-square-roots.png]
{title=The complex number $1+i$ and its two square roots}
{description=Original Argand diagram. Dashed circles show moduli $\sqrt2$ and $2^{1/4}$; the root rays differ by $\pi$ and each squares to the ray of argument $\pi/4$.}
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