Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2004 / Paper 1 / 10

Take the real part of De Moivre's theorem. Put $c=\cos \theta$ and use ${\sin }^2\theta =1-c^2$:

The modulus equation says that $z$ has fixed distance eight from $-1-i$. Thus its locus is a circle of radius eight centered at $(-1,-1)$, with Cartesian equation $(x+1{)}^2+(y+1{)}^2=64$.

Writing $u=X+iY$ gives $X=2x$, $Y=y$. Substitute $x=X/2$, $y=Y$ into the original circle equation. This affine image of a circle is the ellipsecentered at $(-2,-1)$ with horizontal semiaxis sixteen and vertical semiaxis eight. The map covers the whole ellipse as $z$ covers the circle.

The complex translation adds four to the real coordinate and three to the imaginary coordinate. The locus is therefore a circle centered at $(3,2)$ with radius eight:

Multiplication by $i$ rotates the plane counterclockwise through $\pi /2$. The center $3+2i$ maps to $-2+3i$, while distances are preserved. Thus the locus is a circle centered at $(-2,3)$ with radius eight:

For $ab\ne 0$, elimination of the trigonometric parameter givesAs real $\theta$ varies, the whole ellipse is traced, with semiaxes $|a|,|b|$. Signs of $a,b$ affect the traversal direction but not the set. If $a=0$, the locus is the vertical segment $x=0$, $|y|\le |b|$; if $b=0$, it is the horizontal segment $y=0$, $|x|\le |a|$. Both zero give only the origin.

For $ab\ne 0$, the hyperbolic identity gives $x^2/a^2-y^2/b^2=1$. The hyperbolic parametrization of a hyperbola also imposes $x/a=\cosh \theta \ge 1$, so the exact traced locus isIt is only one branch: the right branch for $a>0$ and the left branch for $a<0$. Its vertex is $(a,0)$ and its asymptotes are $y=\pm (b/a)x$. If $a=0$, $b\ne 0$, the entire vertical axis is traced. If $b=0$, $a\ne 0$, the locus is the ray $y=0$, $x/a\ge 1$. Both zero give the origin.