Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2023 / Paper 1 / 11

Direct multiplication, using $i^2=-1$, givesThus the real part is $\boxed{18}$ and the imaginary part is $\boxed{1}$.

Multiply numerator and denominator by the complex conjugate $4-3i$:Thus the real part is $\boxed{6/25}$ and the imaginary part is $\boxed{-17/25}$.

Since $z^2=(x^2-y^2)+2ixy$, equating its real part and imaginary part gives $x^2-y^2=2xy$, orThe quadratic equation in $y$ factors as $[y-(-1+\sqrt{2})x][y-(-1-\sqrt{2})x]=0$. Thus the set is the union of two straight lines through the origin:Their slopes multiply to $-1$, so they are perpendicular. Their angles to the real axis are $\pi /8$ and $-3\pi /8$, modulo $\pi$. The origin is included.

The denominator requires $z\ne 0$. Write $z^2=u+iv$, where $u=x^2-y^2$ and $v=2xy$. Multiplication by the nonzero denominator givesso $u=0$. Conversely, if $u=0$ and $z\ne 0$, then $v\ne 0$ and $v/(iv)=-i$. Thus the required Argand diagram consists exactly ofThese are two perpendicular straight lines, with the origin removed as shown in the figure in part (i).

For real $x$, $\cosh x>0$, so multiplication by the hyperbolic cosine is reversible. The identities $\cosh x-\sinh x=e^{-x}$ and $\mathrm{sech}x=1/\cosh x$ giveThus $e^{-2x}=3$. The exponential function is injective, so the unique real solution is

Set $w=e^z\ne 0$. The hyperbolic cosine identity gives $w+w^{-1}=2i$, hence the quadratic equationLet $L=\ln (1+\sqrt{2})>0$. The first root has modulus $1+\sqrt{2}$ and complex argument $\pi /2$; the second has modulus $\sqrt{2}-1=e^{-L}$ and complex argument $-\pi /2$. Taking all complex logarithm values givesThe two families are distinct because their real parts are respectively $L$ and $-L$. This exhausts the solutions because both roots for $w$ and all possible complex arguments have been included.