Mastering the Tricky Complex Numbers Proof: Tips and Tricks for Success!

lektor
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I recently was confronted by this monstrosity of a question in one of my mock exams.

|Z1 + Z2| ≤ |Z1| + |Z2|

I made a few attempts at it before becoming demoralized with the lack of progress..
|Z^2| was equal to Z1(conjugate)Z1
Hence equaling X^2 + Y^2

However even when expanding into x+iy form etc no avail, help is appreciated.
 
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square both side and see if you notice anything and remember
if A=A then, surely A is less than or equal to A
 
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Geometrically this is just the triangle inequality. It just says that the sum of two sides of a triangle is always greater or equal than the third.

Since you know |z|^2=z*z. Why not write the left side out in this form?
ie: |z+w|^2=(z*+w*)(z+w)
 
I'm sorry but I had to laugh when I read "I recently was confronted by this monstrosity of a question..." only to find the triangle inequality beneathe. lol.
 
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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