Solving Complex Number Inequalities

kukumaluboy
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Homework Statement



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The Attempt at a Solution


-1 < (z-w) /(1-z*w) < 1
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Hi can give clue. I am clueless
 
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kukumaluboy said:
-1 < (z-w) /(1-z*w) < 1
You cannot compare complex numbers with inequalities.

Hint: ##| \frac{a}{b} | = \frac{|a|}{|b|}##
 
|z-w| < |1-z*w| ?? should i square both sides?
 
I would simplify the problem a bit before.

How comfortable are you with substitutions?
I think this problem is solvable without, but it makes life easier if you can simplify it first.
 
Honestly. we have never done these type of questions before. Our exams are always set with questions that we have nvr done before. This is a past year paper haha. Can teach me the one without the substitution ?
 
kukumaluboy said:
Our exams are always set with questions that we have nvr done before.
Well, repeating equations you had before would be pointless, right?

You can write both z and w in terms of real and imaginary part, square both sides and see what you get. It gets much easier if you get rid of one of the four parameters before, however. You can multiply both sides by |z*/z|, that should work.

Edit: Forget the multiplication, it is not as complicated as I expected with the longer approach.
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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