Identify and sketch the region in the complex plane satisfying

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Homework Statement



Identify and sketch the region in the complex plane satisfying

<br /> | \frac{2 z - 1}{z + i} | \geq 1<br />

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The Attempt at a Solution

 
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Geometrically, |z- a| is the distance from z to a as points in the complex plane so I recommend you try to rewrite the given absolute value in that way. It might help to "rationalize the denominator", multiplying both numerator and denominator by z- i.
 
Let z=x+yi. Now simplify your inequality by: |2z-1| greater than or equal to |z+i|.(multiply out |z+i|). sub z=x+yi into this inequality. You will find that it gives a circle (if not its an ellipse, i may have made a quick error in attempting to answer this). The area outside this graph is your answer.
 
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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