Complex Numbers, Square, Conics

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On the Argand diagram, $P$ represents the complex number $z$, and $R$ the number $\frac{1}{z}$ A square $PQRS$ is drawn in the plane with $PR$ as a diagonal If $P$ lies on the circle $|z| = 2$,

(I) Prove that $Q$ will lie on the ellipse whose equation has the form $\frac{x ^ 2}{a ^ 2} + \frac{y ^ 2}{b ^ 2} = 1$;

(II) Hence, specify the numerical values ​​for $a$ and $b$.

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Building upon your start:

The difference between $P$ and the midpoint is: $\frac{3}{4}\cos t+\frac{5}{4}i\sin t$

If we multiple this by $i$ we get a complex number which has been rotated by $90^\circ$. That complex number is: $-\frac{5}{4}\sin t+\frac{3}{4}i\cos t$

Next add this to the midpoint to get $Q$: $\frac{5}{4}\cos t-\frac{5}{4}\sin t+\frac{3}{4}i\cos t+\frac{3}{4}i\sin t$

$$=\frac{5}{4}(\cos t-\sin t)+\frac{3}{4}i(\cos t+\sin t)$$

$$=\frac{5\sqrt{2}}{4}\cos\left(t+\frac{\pi}{4}\right)+\frac{3\sqrt{2}}{4}i\sin\left(t+\frac{\pi}{4}\right)$$

You should recognise this as the parametric form of an ellipse with semi-major radius of $\frac{5\sqrt{2}}{4}$ and semi-minor radius of $\frac{3\sqrt{2}}{4}$.

Hence $a=\frac{5\sqrt{2}}{4}$ and $b=\frac{3\sqrt{2}}{4}$.

Note we could have multiplied by $-i$ for the $90^\circ$ rotation. This would have followed very similar steps and arrived at $Q$:

$$=\frac{5\sqrt{2}}{4}\cos\left(t-\frac{\pi}{4}\right)+\frac{3\sqrt{2}}{4}i\sin\left(t-\frac{\pi}{4}\right)$$

Which would lead to the same $a$ and $b$.