If $w= \frac{3z+i}{i-z}$ show that $\Re(z) \ge0$ implies $\Im(z)\le0$

I have exam after two days and I want answer of the question please help me as fast as you could , thanks everyone . If $w= \frac{3z+i}{i-z}$ show that $\Re(z) \ge0$ implies $\Im(z)\le0$?

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There can be no such implication, with nothing assumed about $w$ other than how to calculate it from $z$. –  coffeemath Jun 18 '13 at 6:42
What should speak against having $z=1+i$? –  Hagen von Eitzen Jun 18 '13 at 6:46
$\Re(z)$ and $\Im(z)$ or $\Re(w)$ and $\Im(w)$? –  Riccardo.Alestra Jun 18 '13 at 7:08
this question is taken from advanced engineering mathematics wylie & barrot 1982 –  user82922 Jun 18 '13 at 20:16

If $z=x+iy$, so that $Re(z)=x$, then a calculation gives $$Im(w)=\frac{-4x}{x^2+(y-1)^2},$$ so that if $Re(z) \ge 0$ you can conclude $Im(w) \le 0.$ So I suspect the problem should have had the output variable $w$ in the imaginary part about which the conclusion of the implication is stated.

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Let $z=x+iy$

$\displaystyle w=\frac{3z+i}{i-z}=\frac{3x+i(1+3y)}{-x+i(1-y)}=\frac{(3x+i(1+3y))(-x-i(1-y))}{x^2+(y-1)^2}=\frac{g(x,y)+i(+3xy-3x+x-3xy)}{x^2+(y-1)^2}=\frac{g(x,y)-i(4x)}{x^2+(y-1)^2}$

Here $g(x,y)$ is the real part of $w$ which is a functiom of $x,y$

So if $x\ge 0$ then $Im(w)\le 0$

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