Show that this function is analytic

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Show that this function is analytic

\left( x + \frac{x}{x^2 + y^2} \right) + i \left( y - \frac{y}{x^2 + y^2} \right)

now... would i substitute x = \frac{z + \overline{z}}{2}
and
y = \frac{z - \overline{z}}{2}

and then see if z or z bar appear exlicitly in the function??
Would that solve it??

Is there an easier way? A less Messy way?
 
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If I recall correctly, a function is analytic if it is differentiable and if it satisfies the Cauchy-Riemann equations, perhaps you should check those?
 
is there another way of doing it... perhaps using the substitutions i suggested above and then checking the domain of the function?
 
stunner5000pt said:
is there another way of doing it... perhaps using the substitutions i suggested above and then checking the domain of the function?

Using the Cauchy-Riemann equations is probably the easier way.
 
stunner5000pt said:
Show that this function is analytic

\left( x + \frac{x}{x^2 + y^2} \right) + i \left( y - \frac{y}{x^2 + y^2} \right)

now... would i substitute x = \frac{z + \overline{z}}{2}
and
y = \frac{z - \overline{z}}{2}

and then see if z or z bar appear exlicitly in the function??
Would that solve it??

Is there an easier way? A less Messy way?

Your equation for y should be over 2i not just 2.
 
stunner5000pt said:
is there another way of doing it... perhaps using the substitutions i suggested above and then checking the domain of the function?
There are complex functions whose domain is C but are nowhere analytic.
 
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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