For real part of $z$ and $w >0$, prove $\left|\frac{z - w}{\bar z + w}\right|< 1$?

Question

So I use the rule $$|z|^2 = z \times\bar z$$ and then in the end I got $$\frac{(|z|^2 + |w|^2 - z\times\bar w-w\times\bar z)}{(|z|^2 + |w|^2 + \bar z\times\bar w+w\times z)}$$

Any help to continue with that what I got?

Thank you for all your answers.

Answer 1

Remember that $|a-b|$ is a distance between (complex) numbers $a$ and $b$ in a complex plane.

We can rewrite it as $$|z-w|<|z-(-\overline{w})|$$

We know that $z$ and $w$ are on the right side of imaginary axis (why?) and $-\overline{w}$ on the left side.

So, clearly $z$ is closer to $w$ than to $-\overline{w}$ so $|z-w|$ is smaller then $|z-(-\overline{w})|$.

and we are done so $

Answer 2

The denominator should be different (change the last term to $+wz$), but everything else, to that point, is fine.

If you want to show this is less than one, then we can equivalently try to show that the numerator is smaller than the denominator. So, we want to show that $$-z\bar{w}-w\bar{z}\leq \bar{z}\bar{w}+w{z}.$$ Let $z=a+bi$ and $w=c+di.$ By assumption, $a,c>0.$ The left-hand side of the desired inequality is $-2ac-2bd,$ and the right is $2ac-2bd.$ Since $a,c>0,$ it follows that $-2ac<2ac.$ So, the left is smaller than the right, as desired.

Answer 3

Well, they told you $Re(z), Re(w) > 0$ so you have been given permission to do use the real and imaginative parts.

So prove $|\frac {(Re(z) - Re(w)) + i(Im(z) - Im(z))}{(Re(z)+R(w)) + i(Im(w) - Im(z)}| < 1$

Which is to say $\frac {\sqrt{(Re(z) -Re(w))^2+ (Im(z) - Im(w))^2}}{\sqrt{(Re(z) + Re(w))^2 + (Im(w) - Im(z))^2}} < 1$ which has nothing to do with complex numbers.

If $a,b>0$ and $c \ge 0$ then

$(a + b)^2 = a^2 + 2ab +b^2 > a^2 - 2ab + b^2 = (a-b)^2$ but both are non-negative.

And $(a+b)^2 + c > (a-b)^2 + c> 0$ so

$\sqrt{(a+b)^2 + c}> \sqrt{(a-b)^2 + c}$ and

$\frac {\sqrt{(a-b)^2 + c}}{\sqrt{(a+b)^2 + c}} < 1$

Let $a = Re(z)$ and $b=Re(w)$ and $c = (Im(w) - Im(z))^2$ and ... we are done.