1. ## Complex number questions

It seems complex numbers has just gotten a new standard of difficulty. I'am unable to post my working at the moment due to not having any start with the Maths Syntax used here. The question as a whole is connected in a manner that I can't solve the later parts getting done with the first part so please help me out.

The variable complex number z is given b:

z = 2cos@ + i(1-2sin@).

where @ takes all values in the interval pi < @ <= pi.

a, Show that lz-il = 2 for all values of @. Hence sketch, in an Argand Diagram, the locus of the point representing z.

b, Prove that the real part of 1/z+2-i is constant for -pi < @ < pi

2. For part a, note that $\displaystyle z - i = 2\cos \left( \Theta \right) - i2\sin \left( \Theta \right)$.

Part b needs a bit more.
$\displaystyle \Re e\left( {\frac{1}{w}} \right) = \frac{{\Re e(w)}} {{\left| w \right|^2 }}$
So $\displaystyle \Re e\left( {\frac{1}{{z + 2 - i}}} \right) = \frac{{2\cos \left( \Theta \right) + 2}}{{8 + 8\cos \left( \Theta \right)}}$

3. a. $\displaystyle z=2(\cos\theta-i\sin\theta)+i$, and given that $\displaystyle |z|=\sqrt{Re^2(z) + Im^2(z)}$ the modulus of z in our case is

$\displaystyle |z-i|=\sqrt{4}\sqrt{\cos^2\theta+\sin^2\theta}=\sqrt{ 4}\sqrt{1}=2.$ Therefore $\displaystyle |z-i|$ is constant at 2.

4. Thank you