What curve does the point describe: 1^(°) z=(1+2it)/(1-it)…

What curve does the point describe: $\ 1^{\circ}\,\, z=\dfrac{1+2it}{1-it}; \quad 2^{\circ}\,\, z=2e^{it}+e^{-it}$, if $t$ is a real parameter?

Part $1^{\circ}$ · $z=\dfrac{1+2it}{1-it}$

Step 1 · Rationalize

$$ z = \frac{(1+2it)(1+it)}{(1-it)(1+it)} = \frac{(1-2t^2)+3it}{1+t^2} $$

so $x=\dfrac{1-2t^2}{1+t^2}$, $\ y=\dfrac{3t}{1+t^2}$.

Step 2 · Combine to eliminate $t$

From $x(1+t^2)=1-2t^2$: $\ (x+2)t^2 = 1-x$, so $t^2=\dfrac{1-x}{x+2}$ and $1+t^2=\dfrac{3}{x+2}$.

Then $$ y^2 = \frac{9t^2}{(1+t^2)^2} = \frac{9\cdot\frac{1-x}{x+2}}{\left(\frac{3}{x+2}\right)^2} = (1-x)(x+2) $$

Step 3 · Simplify

$$ y^2 = 2-x-x^2 \quad\Longrightarrow\quad x^2+x+y^2=2 $$

Step 4 · Complete the square

$$ \left(x+\frac12\right)^2+y^2 = 2+\frac14 = \frac94 $$

$$ \boxed{\left(x+\frac12\right)^2+y^2=\frac94} $$

Answer for $1^{\circ}$

A circle with center $\left(-\frac12,0\right)$ and radius $\frac32$, traced entirely as $t$ ranges over $(-\infty,\infty)$.

Step 5 · Verification

$t=0$: $z=1$, and $\left(1+\frac12\right)^2+0^2=\frac94$ ✓. $\quad t=1$: $z=\frac{1+2i}{1-i}=\frac{-1+3i}{2}$, point $\left(-\frac12,\frac32\right)$, and $0^2+\left(\frac32\right)^2=\frac94$ ✓.

Part $2^{\circ}$ · $z=2e^{it}+e^{-it}$

Step 1 · Expand with Euler's formula

$$ z = 2(\cos t+i\sin t) + (\cos t-i\sin t) = 3\cos t + i\sin t $$

so $x=3\cos t$, $\ y=\sin t$.

Step 2 · Eliminate $t$

$$ \left(\frac{x}{3}\right)^2+y^2 = \cos^2t+\sin^2t = 1 $$

$$ \boxed{\frac{x^2}{9}+y^2=1} $$

Answer for $2^{\circ}$

An ellipse centered at the origin with semi-major axis $a=3$ along the real axis and semi-minor axis $b=1$ along the imaginary axis, vertices at $(\pm3,0)$ and $(0,\pm1)$.

Step 3 · Verification

$t=0$: $z=3$, $\frac99+0=1$ ✓. $\quad t=\frac{\pi}{2}$: $z=2i-i=i$, $\frac09+1=1$ ✓. $\quad t=\pi$: $z=-2-1=-3$, $\frac99+0=1$ ✓.