Question
If $|z|=1$ and $w=\frac{z-1}{z+1}$ where $z \neq-1$ then $\operatorname{Re}($ w)(a) 0(b) $\frac{-1}{|z+1|^{2}}$(c) $\frac{\sqrt{2}}{|z+1|^{2}}$(d) 2
Step 1
We can multiply both sides by $z+1$ to get $w(z+1) = z-1$. Show more…
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If $|z|=1$ and $w=[(z-1) /(z+1)](z \neq-1)$ then $\operatorname{Re}(w)=$ (a) 0 (b) $\left[1 /\left(|z+1|^{2}\right)\right]$ (c) $\left[1 /\left(|z+1|^{3}\right)\right]$ (d) $\left[\sqrt{2} /\left(|\mathrm{z}+1|^{2}\right)\right]$
If $\arg \left(z_{1} z_{2}\right)=0$ and $\left|z_{1}\right|=\left|z_{2}\right|=1$ then (a) $\mathrm{z}_{1}+\mathrm{z}_{2}=0$ (b) $z_{1} z_{2}=1$ (c) $\mathrm{z}_{1}=\overline{\mathrm{z}}_{2}$ (d) $\arg z_{1}=\arg \bar{z}_{2}$
$$ \text { If } z+\sqrt{2}|z+1|+i=0, \text { then }|z|^{2} \text { is } $$
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