Past exam of the mathematics course of the University of Cambridge / 2005 / ia / Paper 1 / 5C (Algebra and Geometry)

The complex exponential function factorizes as $e^{x+iy}=e^x{e}^{iy}=e^x(\cos y+i\sin y)$. HenceThe real exponential multiplies both components; it does not change the argument.

Using $\cos z=(e^{iz}+e^{-iz})/2$ and collecting real and imaginary terms gives the complex cosine identity $\cos (x+iy)=\cos x\cosh y-i\sin x\sinh y$. ThusThe sign follows from the factors $e^{-y}$ and $e^y$ in the two complex exponentials.

For $z\ne 0$, write $z=r{e}^{i\theta }$ with $r=\sqrt{x^2+y^2}>0$. The complex logarithm has values $\ln r+i(\theta +2\pi k)$, $k\in \mathbb{Z}$. ThereforeFor a specified single-valued branch, choose the corresponding continuous determination of the argument. In particular, the principal value uses $\mathrm{Arg}z\in (-\pi ,\pi ]$; it is not represented correctly by $\arctan (y/x)$ in every quadrant. The complex logarithm is undefined at zero.

For $z\ne 0$, rationalizing the denominators gives $1/z=(x-iy)/(x^2+y^2)$ and $1/\overline{z}=(x+iy)/(x^2+y^2)$. Their imaginary parts cancel, soThis also follows because the summands are complex conjugates.

The cubic terms are the expansion of $(z+\overline{z}{)}^3=(2x{)}^3$. Subtracting $\overline{z}=x-iy$ therefore givesHere the expression is a polynomial in $z$ and its complex conjugate, rather than a holomorphic polynomial in $z$ alone.

For the unheaded continuation, set $w=\log z$ in the food equation. The two square roots are $w=\pm i\pi /4$, so exponentiating givesBoth values have $|z-1{|}^2=2-\sqrt{2}<1$ and hence lie strictly inside the disk. Both also have the indicated principal logarithms. Allowing the multivalued complex logarithm produces no additional points: the only permitted values of $w$ are still the two square roots already used.

For drink, the PDF squares only the denominator. Put $z=x+iy$. Its numerator is $(3x+iy)/2$ and its unsquared denominator is $(x+3iy)/2$, which is nonzero unless $z=0$. Multiplying the equation by this denominator squared and equating the real part and imaginary part yieldsthat is $x^2-9y^2=2x$ and $y(1-9x)=0$. If $y=0$, the real equation gives $x=0$ or $x=2$, and zero is excluded. If $y\ne 0$, then $x=1/9$, but the real equation would require $9y^2=-17/81$, impossible for real $y$. ThusIt lies on the allowed disk boundary, and direct substitution gives $3/1^2=3$. The denominator-only square matters: squaring the entire quotient would define a different locus.