How do I prove these complex problems? I need some help on a and b
The unit circle $C$ in the complex plane consists of the complex numbers that have the distance of $1$ from origin. That will say that the complex numbers $z$ so that $|z|=1$.
a) Show that the complex number $z$ is located on $C$ if and only if $zz^\ast=1$.
What I did on a): $|z| \geq 0, zz^\ast=1 \iff |z|^2.$ So $|z|^2=1^2=1$.
b) Assume that $z$ is located in $C.$ Show that one complex number $w$ is located on the tangent to $C$ through $z$ only and only if $w=z(1+it)$ for a real number $t$. Show that this is equivalent to $w+z^2w^\ast=2z$.
b) $w-z=z(e^{\pi i/2}\cdot t) =zti$. I don't know what to do anymore.
 A: a) $|z| = 1 \iff  z \bar{z}=1$ To proof it we will proof every side of the $\iff$ 


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*a1) $ |z| = 1 \implies z\bar{z}=1$
Proof: Let $z=a+ib$ and $ \bar{z}=a-ib$ then $|z|=\sqrt{a^2+b^2} =1$ so $a^2+b^2=1 $ then   $\bar{z}=a-ib$ so $z\bar{z}=(a+ib)(a-ib)=a^2+b^2 = 1$

*a2) $ z\bar{z}=1 \implies |z| = 1$ 
Proof: Let $z=a+ib$ then $\bar{z}=a-ib$ so
$z\bar{z}=(a+ib)(a-ib)=a^2+b^2=|z| = 1$
b) Imagine rather than working with complex numbers you were working with Euclidean 2d vectors. Same rules, same thing. Cosine product ($\cdot$) is your friend, two vectors are orthogonal if their cosine product is 0, since $cos(90)=0$. So they are on the tangent if and only if they are orthogonal. The proof then is as follows, per the definition: 
Imagine $z=a+bi$ then $w= z(1+ti) = a +ati + bi  -bt =  a -bt + i(b + at)$ 
The vector $w-z =  a -bt - a + i(at) = -bt + ati$ then $(w-z)\cdot z= -abt +atb = 0$ 
This can be used in booth sense to demonstrate what you need. It is much easier than writing it here. 
Does it help? 
A: Part (a). It seems to me you have the right proof, but it's not expressed well. For example, the part "$z\bar{z}=1 \Longleftrightarrow |z|^2$" doesn't make much sense, because "$|z|^2$" on the right is only an expression, not a statement. (And it's not my fault, because you confirmed that my edits didn't ruin your work.☺) Also, using the word "so" in the end makes it a sound as a one-way implication, not a two-way equivalence. It may seem like a little detail, but written like this your proof goes in one direction only.
Assuming you already know and are allowed to use the fact that $z\bar{z}=|z|^2$, a proof is pretty straightforward, and you pretty much have it:
$$z\bar{z}=1 \Longleftrightarrow |z|^2=1 \Longleftrightarrow |z|=1,$$
where the last equivalence is true precisely because $|z|\ge0$, as you said.
Part (b). The key is that line tangent to a circle is perpendicular to the radius going to the point of tangency, and I think you have that too. If $w$ lies on the tangent line to the circle $C$ at the point $z$, then $w-z$ is perpendicular to the radius from the origin to $z$. But so is $ze^{i\pi/2}$, since multiplication by $e^{i\pi/2}$ is rotation by $\pi/2$ (counterclockwise). So $w-z$ and $ze^{i\pi/2}$ have the same or opposite direction. Think of them as vectors — they are parallel. But since they don't have to have the same length, they are multiples of each other by some real coefficient $t$, so $w-z=t\cdot ze^{i\pi/2}$, or $w-z=zti$, as desired.
Now, that that's been done, we know that $w=z(1+it)$. Then $\bar{w}=\bar{z}(1-it)$, and
$$w+z^2\bar{w}=z(1+it)+z^2\bar{z}(1-it)=z(1+it)+z\cdot \underbrace{z\bar{z}}_{|z|^2=1}\cdot(1-it)=z(1+it)+z(1-it),$$
giving the desired result.
