How to prove |z-i|=2 with $z – i = 2\cos\theta – 2i\sin\theta $

complex numbers

I have the following question. It's basically my first day doing complex numbers, so I am absolutely lost here. enter image description 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.

Best Answer

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?
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