Chapter 17

\(x^{2}-y^{2}-4 x+(-2 x y-4 y) i=0+0 i\) implies \(x^{2}-y^{2}-4 x=0\) and \(y(-2 x-4)=0 .\) If \(y=0\) then \(x(x-4)=0\) and so \(z=0\) and \(z=4 .\) If \(-2 x-4=0\) or \(x=-2\) then \(12-y^{2}=0\) or \(y=\pm 2 \sqrt{3} .\) This gives \(z=-2+2 \sqrt{3} i\) and \(z=-2-2 \sqrt{3} i\)

\(|10+8 i|=\sqrt{164}\) and \(|11-6 i|=\sqrt{157}\). Hence \(11-6 i\) is closer to the origin.

By the quadratic formula, \(e^{z}=-\frac{1}{2}+\frac{\sqrt{3}}{2} i\) or \(e^{z}=-\frac{1}{2}-\frac{\sqrt{3}}{2} i .\) Hence $$z=\ln \left(-\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)=\left(\frac{2 \pi}{3}+2 n \pi\right) i \quad \text { or } \quad z=\ln \left(-\frac{1}{2}-\frac{\sqrt{3}}{2} i\right)=\left(\frac{4 \pi}{3}+2 n \pi\right) i$$

We have $$(\cos \theta+i \sin \theta)^{3}=\cos 3 \theta+i \sin 3 \theta$$ Also $$\begin{aligned} (\cos \theta+i \sin \theta)^{3} &=\cos ^{3} \theta+3 \cos ^{2} \theta(i \sin \theta)+3 \cos \theta(i \sin \theta)^{2}+(i \sin \theta)^{3} \\ &=\cos ^{3} \theta-3 \cos \theta \sin ^{2} \theta+\left(3 \cos ^{2} \theta \sin \theta-\sin ^{3} \theta\right) i \end{aligned}$$ Equating real and imaginary parts gives $$\cos 3 \theta=\cos ^{3} \theta-3 \cos \theta \sin ^{2} \theta, \quad \sin 3 \theta=3 \cos ^{2} \theta \sin \theta-\sin ^{3} \theta$$

$$\left|\frac{1}{2}-\frac{1}{4} i\right|=\frac{\sqrt{5}}{4} \text { and }\left|\frac{2}{3}+\frac{1}{6} i\right|=\frac{\sqrt{17}}{6}, \text { since } \frac{\sqrt{5}}{4}<\frac{\sqrt{17}}{6}, \frac{1}{2}-\frac{1}{4} i \text { is closer to the origin. }$$

We have $$\lim _{\Delta z \rightarrow 0} \frac{\overline{z+\Delta z}-\bar{z}}{\Delta z}=\lim _{\Delta z \rightarrow 0} \frac{\overline{\Delta z}}{\Delta z}.$$ If we let \(\Delta z \rightarrow 0\) along a horizontal line then \(\Delta z=\Delta x, \overline{\Delta z}=\Delta x,\) and $$\lim _{\Delta z \rightarrow 0} \frac{\overline{\Delta z}}{\Delta z}=\lim _{\Delta x \rightarrow 0} \frac{\Delta x}{\Delta x}=1.$$ If we let \(\Delta z \rightarrow 0\) along a vertical line then \(\Delta z=i \Delta y, \overline{\Delta z}=-i \Delta y,\) and $$\lim _{\Delta z \rightarrow 0} \frac{\overline{\Delta z}}{\Delta z}=\lim _{\Delta y \rightarrow 0} \frac{-i \Delta y}{i \Delta y}=-1.$$ Since these two limits are not equal, \(f(z)=\bar{z}\) cannot be differentiable at any \(z\).

$$(-i)^{4 i}=e^{4 i \ln (-i)}=e^{4 i\left[\log _{e} 1+i\left(-\frac{\pi}{2}+2 n \pi\right)\right]}=e^{(2-8 n) \pi}$$

We have \(f^{\prime}(z)=\lim _{\Delta z \rightarrow 0} \frac{(z+\Delta z)(\bar{z}+\overline{\Delta z})-z \bar{z}}{\Delta z}=\lim _{\Delta z \rightarrow 0}\left(\bar{z}+z \frac{\overline{\Delta z}}{\Delta z}+\overline{\Delta z}\right).\) If \(z=0,\) then the above limit becomes $$f^{\prime}(0)=\lim _{\Delta z \rightarrow 0} \overline{\Delta z}=0.$$ If \(z \neq 0\) then we first let \(\Delta z \rightarrow 0\) along a horizontal line so that \(\Delta z=\Delta x\) and \(\overline{\Delta z}=\Delta x .\) Thus, $$f^{\prime}(z)=\lim _{\Delta z \rightarrow 0}\left(\bar{z}+z \frac{\Delta x}{\Delta x}+\Delta x\right)=\bar{z}+z.$$ Next we let \(\Delta z \rightarrow 0\) along a vertical line so that \(\Delta z=i \Delta y, \overline{\Delta z}=-i \Delta y .\) Thus $$f^{\prime}(z)=\lim _{\Delta y \rightarrow 0}\left(\bar{z}+z \frac{-i \Delta y}{i \Delta y}+i \Delta y\right)=\bar{z}-z.$$ We must have \(\bar{z}+z=\bar{z}-z\) which implies \(z=0\).This is a contradiction to the assumption that \(z \neq 0 .\) Hence \(f(z)=|z|^{2}\) is differentiable only at \(z=0.\)

(a) If we take \(\arg \left(z_{1}\right)=\pi\) and \(\arg \left(z_{2}\right)=\pi / 2\) then \(\arg \left(z_{1}\right)+\arg \left(z_{2}\right)=3 \pi / 2\) is an argument of the product \(z_{1} z_{2}=-5 i .\) With these same arguments we see that \(\arg \left(z_{1}\right)-\arg \left(z_{2}\right)=\pi / 2\) is an argument of the quotient \(z_{1} / z_{2}=\frac{1}{5} i\) (b) If we take \(\arg \left(z_{1}\right)=\pi\) and \(\arg \left(z_{2}\right)=-\pi / 2\) then \(\arg \left(z_{1}\right)+\arg \left(z_{2}\right)=\pi / 2\) is an argument of the product \(z_{1} z_{2}=5 i .\) With these same arguments we see that \(\arg \left(z_{1}\right)-\arg \left(z_{2}\right)=3 \pi / 2\) is an argument of the quotient \(z_{1} / z_{2}=-\frac{1}{5} i\)

Each linear equation in the system \(\frac{d x}{d t}=2 x, \quad \frac{d y}{d t}=2 y\) can be solved directly. We obtain \(x(t)=c_{1} e^{2 t}\) and \(y(t)=c_{2} e^{2 t}.\)

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 \(y=(x-1)^{2}\) 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.

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 \\]

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).