Any ideas for these two complex numbers exercises?

Question

Let $z\in\mathbb{C}$ such that $\left|z-a\right|=\sqrt{a^2-b^2}$, $a>b>0$. Calculate $\left|\frac{b-z}{b+z}\right|$.

I tried to amplify with the conjugate and try to work something out… or square the first relation but couldn't get anywhere…

Let $a\in\mathbb{C}$. Calculate the value of the expression $$E=\left|a+\frac12\right|^2+i\left|a+i\frac12\right|^2-(1+i)\left|a\right|^2-\frac14(1+i).$$

Here give your answer in function of $a$.

Answer 1

I did not understand your question properly but am writing the answer for the first question

Let $z=x+yi$, hence from given given condition $$(x-a) ^2+y^2=a^2-b^2$$ $$\Rightarrow x^2+y^2+b^2=2ax$$

We need to find $\vert \frac {b-z}{b+z}\vert=\frac {\vert b-z\vert}{\vert b+z\vert}$

Hence we get $$\vert \frac {b-z}{b+z}\vert=\sqrt {\frac {(b-x) ^2+y^2}{(b+x)^2+y^2}}$$ Using the above relation we simplify the rest as $$\sqrt {\frac {(b-x) ^2+y^2}{(b+x)^2+y^2}}=\sqrt {\frac {a-b}{a+b}}$$

Similarly for second question

Let $a=x+yi$ And substitute it the given expression for E we get $$E= (x^2+\frac 14+y^2+x-x^2-y^2-\frac 14) +i(x^2+\frac 14+y+y^2-x^2-y^2-\frac 14) $$ Which simplifies to $x+yi$

Hence $E=a$

Answer 2

Let a = b + ic

then E = (b + (1 / 2 ))^2 + c^2 + i * (b^2 + ( c + (1 / 2 ))^2 - ( 1 + i )(b^2 + c^2 ) - ( 1 / 4 ) ( 1 + i )

E = b^2 + b + ( 1 / 4 ) + c^2 - b^2 - c^2 - 1 / 4 + i*[ ( b^2 + ( c + ( 1 / 2 )^2 )^2 - b^2 - c^2 - ( 1 / 4 ) ]

E = b + i * ( b^4 + 2b^2*(c + 0.5)^2 + c^2 + c + 1/4 - b^2 - c^2 - 1/ 4 )

E = b + i * (b^4 + [2(c + 0.5 )^2 - 1 ]*b^2 + c )