Because an argument of the product \((z)(z)\) is twice an argument of \(z\text{,}\) we say that \(f\) doubles angles at the origin. Points that lie on the ray \(r>0\text{,}\)\(\theta =\alpha\) are mapped onto points that lie on the ray \(\rho >0\text{,}\)\(\phi =2\alpha\text{.}\) If we now restrict the domain of \(w=f(z) =z^2\) to the region
We leave as an exercise to show that \(f\) and \(g\) satisfy Equations (2.1.3) and thus are inverses of each other that map the set \(A\) one-to-one and onto the set \(B\) and the set \(B\) one-to-one and onto the set \(A\text{,}\) respectively. Figure 2.2.2 illustrates this relationship.
Show that the transformation \(w=f(z) =z^2\text{,}\) for \(z\ne 0\text{,}\) usually maps vertical and horizontal lines onto parabolas and use this fact to find the image of the rectangle \(\{(x, y) : 0\lt x\lt a, \; 0\lt y\lt b\}\text{.}\)
Using Equations (2.2.4), we determine that the vertical line \(x=a\) is mapped onto the set of points given by the equations \(u=a^2-y^2\) and \(v=2ay\text{.}\) If \(a \ne 0\text{,}\) then \(y=\frac{v}{2a}\) and
Equation (2.2.5) represents a parabola with vertex at \(a^2\text{,}\) oriented horizontally, and opening to the left. If \(a>0\text{,}\) the set \(\{(u, v) : u=a^2-y^2, \; v=2ay\}\) has \(v>0\) precisely when \(y>0\text{,}\) so the part of the line \(x=a\) lying above the \(x\) axis is mapped to the top half of the parabola.
The horizontal line \(y=b\) is mapped onto the parabola given by the equations \(u=x^2-b^2\) and \(v=2xb\text{.}\) If \(b \ne 0\text{,}\) then as before we get
\begin{equation}
u = -b^2 + \frac{v^2}{4b^2}\text{.}\tag{2.2.6}
\end{equation}
Equation (2.2.6) represents a parabola with vertex at \(-b^2\text{,}\) oriented horizontally and opening to the right. If \(b>0\text{,}\) the part of the line \(y=b\) to the right of the \(y\) axis is mapped to the top half of the parabola because the set \(\{(u, v) : u = x^2-b^2, \; v=2bx\}\) has \(v>0\) precisely when \(x>0\text{.}\)
Quadrant I is mapped onto quadrants I and II by \(w=z^2\text{,}\) so the rectangle \(0\lt x\lt a\text{,}\)\(0\lt y\lt b\) is mapped onto the region bounded by the top halves of the parabolas given by Equations (2.2.5) and (2.2.6) and the \(u\) axis. The vertices 0, \(a\text{,}\)\(a+ib\text{,}\) and \(ib\) of the rectangle are mapped onto the four points 0, \(a^2\text{,}\)\(a^2-b^2+i2ab\text{,}\) and \(-b^2\text{,}\) respectively, as indicated in Figure 2.2.4.
Finally, we can verify that the vertical line \(x=0, \, y \ne 0\) is mapped to \(\{(-y^2, 0) : y \ne 0\}\text{.}\) This is simply the set of negative real numbers. Likewise, the horizontal line \(y=0, \, x \ne 0\) is mapped to the set \(\{(x^2, 0) : x \ne 0\}\text{,}\) which is the set of positive real numbers.
where \(-\pi \lt \theta \le \pi\text{?}\) If we use polar coordinates for \(w=\rho e^{i\phi}\) in the \(w\) plane, we can represent this mapping by the system
\begin{equation}
\rho =r^\frac{1}{2} \text{ and } \phi =\frac{\theta}{2}\text{.}\tag{2.2.7}
\end{equation}
Equations (2.2.7) indicate that the argument of \(f(z)\) is half the argument of \(z\) and that the modulus of \(f(z)\) is the square root of the modulus of \(z\text{.}\) Points that lie on the ray \(r>0\text{,}\)\(\theta =\alpha\) are mapped onto the ray \(\rho >0, \, \phi =\frac{\alpha}{2}\text{.}\) The image of the \(z\) plane (with the point \(z=0\) deleted) consists of the right half-plane \(\mathrm{Re}(w) > 0\) together with the positive \(v\) axis. The mapping is shown in Figure 2.2.5
We can use knowledge of the inverse mapping \(z=w^2\) to get further insight into how the mapping \(w=z^{\frac{1}{2}}\) acts on rectangles. If we let \(z=x+iy \ne 0\text{,}\) then
and we note that the point \(z=x+iy\) in the \(z\) plane is related to the point \(w=u+iv=z^{\frac{1}{2}}\) in the \(w\) plane by the system of equations
\begin{equation}
x=u^2-v^2 \text{ and } y=2uv\text{.}\tag{2.2.8}
\end{equation}
Let \(a>0\text{.}\) Equations (2.2.8) map the right half-plane given by \(\mathrm{Re}(z) >a\) (i.e., \(x>a\)) onto the region in the right half-plane satisfying \(u^2-v^2>a\) and lying to the right of the hyperbola \(u^2-v^2=a\text{.}\) If \(b>0\text{,}\) Equations (2.2.8) map the upper half-plane \(\mathrm{Im}(z) >b\) (i.e., \(y>b\)) onto the region in quadrant I satisfying \(2uv>b\) and lying above the hyperbola \(2uv=b\text{.}\) This situation is illustrated in Figure 2.2.7. We leave as an exercise the investigation of what happens when \(a=0\) or \(b=0\text{.}\)
We can easily extend what we’ve done to integer powers greater than 2. We begin by letting \(n\) be a positive integer, considering the function \(w=f(z) =z^n\text{,}\) for \(z=re^{i\theta} \ne 0\text{,}\) and then expressing it in the polar coordinate form
The image of the ray \(r>0, \, \theta =\alpha\) is the ray \(\rho >0\text{,}\)\(\phi =n\alpha\text{,}\) and the angles at the origin are increased by the factor \(n\text{.}\) The functions \(\cos n\theta\) and \(\sin n\theta\) are periodic with period \(2\pi /n\text{,}\) so \(f\) is in general an \(n\)-to-one function; that is, \(n\) points in the \(z\) plane are mapped onto each non-zero point in the \(w\) plane.
which consists of all points in the \(w\) plane except the point \(w=0\text{.}\) The inverse mapping of \(f\text{,}\) which we denote \(g\text{,}\) is then
We leave as an exercise to show that \(f\) and \(g\) are inverses of each other that map the set \(E\) one-to-one and onto the set \(F\)\ and the set \(F\) one-to-one and onto the set \(E\text{,}\) respectively. Figure 2.2.9 illustrates this relationship.
Using Equations (2.2.4) we see that, if \(A=\{(x,y):y=1\}\text{,}\) then \(f(A)=\{(u,v):u=x^2-1,\;v=2x\} = \{(u,v):u=\frac{v^2}{4}-1\}\text{.}\) The region in the upper half plane \(\mathrm{Im}(w)>0\) that lies between the parabolas \(u=4-\frac{v^2}{16}\) and \(u=\frac{v^2}{4}-1\text{.}\)
The point \((x,y)\) in the \(xy\) plane is mapped to the point \((u,v)=(x^2-y^2, \, 2xy)\text{.}\) For any \(x, \; u=x^2-\frac{v^2}{4x^2}\text{.}\) If \(x=1\) then \(u=1-\frac{v^2}{4}\text{.}\)If \(x=2\) then \(u=4-\frac{v^2}{16}\text{.}\) Your only remaining task is to show that the strip \(\{(x,y):1\lt x\lt 2\}\) is mapped between these two parabolas.
The infinite strip \(\{(u,v):1\lt v\lt 2\}\text{,}\) which is the region in the \(uv\) plane between \(v=i\) and \(v=2i\text{.}\) Show the details in a manner similar to the answer for part a.
See also problem 2. The fallacy lies in the assumption implicit in the second equality that \(\sqrt{z_1z_2}=\sqrt{z_1}\sqrt{z_2}\) for all complex numbers \(z_1\) and \(z_2\text{.}\) Assuming the principal square root is used, then \(\sqrt{z_1z_2} = |z_1z_2|^{\frac{1}{2}} e^{i\frac{{\mathrm{Arg}}(z_1z_2)}{2}}\text{.}\) This quantity will equal \(\sqrt{z_1}\sqrt{z_2} = |z_1|^{\frac{1}{2}} e^{i\frac{{\mathrm{Arg}}(z_1)}{2}}|z_2|^{\frac{1}{2}}e^{i\frac{{\mathrm{Arg}}(z_2)}{2}}\) precisely when \({\mathrm{Arg}}(z_1z_2) ={\mathrm{Arg}}(z_1) +{\mathrm{Arg}}(z_2)\)—explain! The latter equality is plainly false when \(z_1=z_2=-1\text{.}\) (Again, explain.) To give a very thorough answer to this problem, you should state precisely when the last equality is true, and prove your assertion.
Show that the functions \(f(z) =z^2\) and \(g(w) =w^{\frac{1}{2}} = |w|^{\frac{1}{2}}e^{i\frac{{\mathrm{Arg}}(w)}{2}}\) with domains given by Equations (2.2.1) and (2.2.2), respectively, satisfy Equations (2.1.3). Thus, \(f\) and \(g\) are inverses of each other that map the shaded regions in Figure 2.2.5 one-to-one and onto each other.
The right half plane given by \(\mathrm{Re}(z)>0\) is mapped onto the region in the right half plane satisfying \(u^2-v^2>0\) and lies to the right of \(u^2-v^2=0\text{.}\) This is the region between the lines \(u=v\) and \(u=-v\) in the right half of the \(w\) plane. A similar analysis can be applied to the case where \(b=0\text{.}\)
