Analyticity of a function in $x$ and $y$, without employing the Cauchy-Riemann eqns

Exercise from Saff & Snider's Complex Analysis:

How to determine the analyticity of this function, without using the Cauchy-Riemann equations? I tried to work from first principles (taking the limit at $h=0$ of the difference quotient) but because the function is given in $x$ and $y$, I can't figure out how to apply that:

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

P.S. Certainly, I can see that the denominators are $|z|^2$, though I'm not sure that introducing the modulus would be helpful.

• Maybe they expect you to recognize $z+1/z$? – 75064 May 19 '13 at 6:02
• Note that for $z=x+iy$ and $\bar{z}=x-iy$, $|z|^2=z\bar{z}$. – 23rd May 19 '13 at 6:10
• @75064 Thanks, neat observation! But I wonder how I could've recognised it on my own? – Ryan May 19 '13 at 6:35
• By practicing the arithmetics of complex numbers. Having calculated $1/(3+4i)$ and other such things, one becomes familiar with the reciprocals of complex numbers, and thus able to recognize them. – 75064 May 19 '13 at 6:37
• If that expression is called $f(x,y)$, you can form the difference quotient $\dfrac{f(x+h,y+k)-f(x,y)}{h+ik}$, then try to simplify and show that the limit as $(h,k)\to (0,0)$ exists. – Jonas Meyer May 19 '13 at 6:46

As said in the comments: while in principle one could try to show that $$\lim_{(h,k)\to (0,0)} \dfrac{f(x+h,y+k)-f(x,y)}{h+ik}$$ exists without switching to complex notation, the fraction with $x^2+y^2$ in denominator will make this computation messy. It is far easier to recognize that $$\frac{x-iy}{x^2+y^2} = \frac{\bar z}{|z|^2 } = \frac{1}{z}$$ Then you can use complex arithmetics: $$\frac{1}{z+h}-\frac{1}{z} = - \frac{h}{z(z+h)}$$ hence the derivative is $-1/z^2$. (The linear term, which is $z$, is not a problem with either approach).