Solve the complex analysis problem

chwala
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The problem is attached

regards,

chwala ken
 

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Nope dude, that's incorrect. x and y are wrong. Try forming a real-valued denominator and then check again what x and y are.
 
i do not get it ? A complex number has a Real part and an Imaginary part. The idea is to breakdown the expression in the form x+iy so i do not understand why you are talking of real valued denominator. Kindly expound by looking at the problem and check where i went wrong.
 
Suppose z=1/(a+i*b)
Then z not equal to 1/a + i*1/b
Why? Basic fraction calculus. You cannot "split up" a fraction among its denominator, which is what you're intending. You can split it up among the nominator though, so you should make the denominator real-valued by expanding with its complex conjugate, that is
z=(a-i*b)/((a+i*b)*(a-i*b))=(a-i*b)/(a^2+b^2)
so your final expression would be
a/(a^2+b^2)-i*b/(a^2+b^2)

If you don't understand the above, my advice would be to real a bit about basics of complex numbers on wikipedia (no offense)
 
Ha..,Thanks Susskind for the insight.
 
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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