Consider the harmonic function $$u(x,y)=1-y+\frac{x}{x^2+y^2}$$ on the upper half plane $y>0$.
What is the corresponding harmonic function on the first quadrant $x>0$, $y>0$, under the transformation $z\mapsto z^2$?
Do I have to use Cauchy-Reimann equations to find the corresponding harmonic function? Any help would be appreciated. Thanks.
If I understand correctly, all that you should do is plug in $z^2$ instead of $z$. No need for the Cauchy-Riemann equations.
First, write $u$ in terms of $z$: $$ u(z)=1- \operatorname{Im} z + \operatorname{Re} \frac{1}{z} $$ Then plug in $z^2$: $$ u(z^2)=1- \operatorname{Im} z^2 + \operatorname{Re} \frac{1}{z^2} $$ Optionally, return to $z=x+iy$ notation: $$ u(z^2)=1- \operatorname{Im} z^2 + \frac{\operatorname{Re} z^2}{|z|^4} =1-2xy+\frac{x^2-y^2}{(x^2+y^2)^2} $$