Past exam of the mathematics course of the University of Cambridge / 2014 / ia / Paper 1 / 1B (Vectors and Matrices)

Squaring first avoids unnecessary expansion: $(2+2i{)}^2=8i$, and squaring again gives the fourth power $\boxed{z^4=-64}$. In the polar form of a complex number, $z=2\sqrt{2}{e}^{i\pi /4}$; multiplying its complex argument by four gives the same negative-real result.

Write a nonzero complex number as $w=r{e}^{i\theta }$. Matching modulus and complex argument gives $r^4=2\sqrt{2}$ and $4\theta =\pi /4+2\pi k$. Thus the four distinct fourth roots areThese angles differ by $\pi /2$, so the values are distinct. Every further integer $k$ repeats one of them; this is multiplication by the fourth roots of unity.

The exponent in the PDF is $2z^2$. A complex exponential equals one precisely when its exponent is an integer multiple of $2\pi i$: its real part must be zero, and its imaginary part must be a full rotation. Therefore $z^2=\pi ik$ for $k\in \mathbb{Z}$. Taking both square roots yields all solutionsThe first nonzero family corresponds to positive exponent index, and the second to negative index. Both signs are required.

For a real line $ux+vy+w=0$, with $(u,v)\ne (0,0)$, choose $B=(u-iv)/2$ and $C=w$. Direct multiplication gives $Bz+\overline{B}\overline{z}=ux+vy$, so every line has the required form with $B\ne 0$.

For a circle of centre $c$ and radius $\rho >0$, expand its modulus equation:Thus the required circle parameters are $\boxed{B=-c,C=|c{|}^2-{\rho }^2}$. Conversely, completing the square in the circle form gives $|z+B{|}^2=|B{|}^2-C$, which describes a genuine circle when $|B{|}^2-C>0$. The complex conjugate terms make both equations real-valued.