91Ó°ÊÓ

The equations in the system \(\frac{d x}{d t}=\frac{x}{x^{2}+y^{2}}, \quad \frac{d y}{d t}=\frac{y}{x^{2}+y^{2}}\) can be divided to give \(\frac{d y}{d x}=\frac{y}{x}\).By separation of variables we obtain \(y=c x.\)

If \(y=\frac{1}{2} x^{2}\) the equations \(u=x^{2}-y^{2}, v=2 x y\) give \(u=x^{2}-\frac{1}{4} x^{4}, v=x^{3}\).With the aid of a computer, the graph of these parametric equations is shown.

If \(z_{1}=-i\) and \(z_{2}=i\) then \\[ \operatorname{Ln}\left(z_{1} / z_{2}\right)=\operatorname{Ln}(-1)=\pi i, \quad \text { whereas } \quad \operatorname{Ln} z_{1}-\operatorname{Ln} z_{2}=-\frac{\pi}{2} i-\frac{\pi}{2} i=-\pi i \\]

If \(y=(x-1)^{2}\) the equations \(u=x^{2}-y^{2}, v=2 x y\) give \(u=x^{2}-(x-1)^{4}, v=2 x(x-1)^{2} .\) With the aid of a computer the graph of these parametric equations is shown.

Since \(|z|=\sqrt{x^{2}+y^{2}}\) and \(\operatorname{Arg} z=\tan ^{-1} \frac{y}{x}\) for \(x>0\) we have \\[ \operatorname{Ln} z=\log _{e}|z|+i \operatorname{Arg} z=\log _{e}\left(x^{2}+y^{2}\right)^{1 / 2}+i \tan ^{-1} \frac{y}{x}=\frac{1}{2} \log _{e}\left(x^{2}+y^{2}\right)+i \tan ^{-1} \frac{y}{x} \\]

(a) \(u=\log _{e}\left(x^{2}+y^{2}\right) ; \quad \frac{\partial^{2} u}{\partial x^{2}}=\frac{2\left(y^{2}-x^{2}\right)}{\left(x^{2}+y^{2}\right)^{2}}, \quad \frac{\partial^{2} u}{\partial y^{2}}=\frac{2\left(x^{2}-y^{2}\right)}{\left(x^{2}+y^{2}\right)^{2}}\) since \(\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0\) the function \(u\) is harmonic in any domain not containing the point (0,0). (b) \(v=\tan ^{-1} \frac{y}{x} ; \quad \frac{\partial^{2} v}{\partial x^{2}}=\frac{2 x y}{\left(x^{2}+y^{2}\right)^{2}}, \quad \frac{\partial^{2} v}{\partial y^{2}}=\frac{-2 x y}{\left(x^{2}+y^{2}\right)}\) since \(\frac{\partial^{2} v}{\partial x^{2}}+\frac{\partial^{2} v}{\partial y^{2}}=0\) the function \(v\) is harmonic in any domain not containing the point (0,0).