How do I prove |z-i|=2 with $z - i = 2\cos\theta - 2i\sin\theta $?

Question

I have the following question. It's basically my first day doing complex numbers, so I am absolutely lost here. I have read that the modulus-arg form is $$ z = r(\cos\theta + i \sin\theta)$$ Now, in this case, I tried expanding the equation given (I'm only on part i right now) and got:

$$z - i = 2\cos\theta - 2i\sin\theta $$ What do I do now? Yes, I can factor the 2 out, but my issue is that I was told that the value of r and the signs of cos and sin must be positive for the mod-arg form. I'm not sure what to do.

If $z=x+iy,|z|=\sqrt{x^2+y^2}$ – Shubham Johri Oct 26 '20 at 19:15 — Shubham Johri, Oct 26 '20 at 19:15
$\sin\theta=\sin(-\theta)$ and $\cos\theta=\cos(-\theta)$. – Bernard Oct 26 '20 at 19:34 — Bernard, Oct 26 '20 at 19:34
$z-i=2(e^{-i\theta}) $ $\Rightarrow $$ |z-i|=2$ – Oct 26 '20 at 20:28 — , Oct 26 '20 at 20:28

score 1 · Answer 1 · answered Oct 26 '20 at 19:15

1

By definition we have that

$$|x+iy|^2=x^2+y^2$$

then

$$|z-i|^2= (2\cos \theta)^2+(-2\sin \theta)^2$$

then use $\cos^2 \theta + \sin^2 \theta =1$.

answered Oct 26 '20 at 19:15

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Alessio K · Accepted Answer · 2020-10-26T20:36:12.293

Hints:

For $z=x+iy$ we have that $|z|=\sqrt{x^2+y^2}$ and use the identity $\sin^2(\theta)+\cos^2(\theta)=1$
$|z-i|=2$ is the set of all points, $z$, in the Argand diagram with a distance of $2$ from $i$. Write $z=x+iy$ and evaluate $|z-i|=2$ to find the equation of the circle.
$\frac{1}{z+2-i}=\frac{1}{2\cos(\theta)+2-2i\sin(\theta)}=\frac{2\cos(\theta)+2+2i\sin(\theta)}{4(\cos(\theta)+1)^2+4\sin^2(\theta)}=\frac{2(\cos(\theta)+1)+2i\sin(\theta)}{8(1+\cos(\theta))}$ by rationalising the denominator. So what is the real part?

How do I prove |z-i|=2 with $z - i = 2\cos\theta - 2i\sin\theta $?

2 Answers2