2.1 Complex function

1. For $z = x + iy$, $w = u + iv$, calculate $w = z^2$. Then

$$w = z^2 = (x + iy)^2 = x^2 - y^2 + i(2xy) = u + iv$$

Thus, $u = x^2 - y^2$, $v = 2xy$. Next we solve this for $x,y$.

Since $v = 2xy$, $y = \frac{v}{2x}$. Put this into $u = x^2 - y^2$. Then

$$u = x^2 - (\frac{v}{2x})^2 = x^2 - \frac{v^2}{4x^2}$$

Multiply both sides by $4x^2$. Then

$$4x^4 - 4x^2 u - v^2 = 0$$

This can be thought of quadratic equation in $x^2$. Thus let $x^2 = X$. Then. by the formula of quadratic equation, we have

$$x^2 = \frac{2u \pm \sqrt{4u^2 + 4v^2}}{4} = \frac{u \pm \sqrt{u^2 + v^2}}{2}$$

Note that $x$ is a real part. Thne

$$x^2 = \frac{u + \sqrt{u^2 + v^2}}{2}$$

Thus we have

$$x = \pm \sqrt{\frac{u + \sqrt{u^2 + v^2}}{2}}$$

Next find $y$. Since $u = x^2 - y^2$, we have

$$y^{2} = x^2 - u = \frac{u + \sqrt{u^2 + v^2}}{2} - u = \frac{-u + \sqrt{u^2 + v^2}}{2}$$

Thus

$$y = \pm \sqrt{\frac{-u + \sqrt{u^2 + v^2}}{2}}$$

Next, consider what kind of curve the straight line $y = \pm b (b> 0)$ parallel to the real axis of the $z$ plane is mapped. By the above equation, $y = \pm b$ satisfies

$$y = \pm b = \pm \sqrt{\frac{-u + \sqrt{u^2 + v^2}}{2}}$$

Thus,

$$b^2 = \frac{-u + \sqrt{u^2 + v^2}}{2}$$

and it is mapped to the parabola $v^2 = 4b^2(b^2 + u)$

Similarly, a straight line parallel to the imaginary axis of the $z$ plane $x = \pm a (a > 0)$ satifies

$$x = \pm a = \pm \sqrt{\frac{u + \sqrt{u^2 + v^2}}{2}}$$

Thus

$$a^2 = \frac{u + \sqrt{u^2 + v^2}}{2}$$

and it is mapped to parabola $v^2 = 4a^2(a^2 - u)$.

2.

(a) $z = x + iy$, $w = u + iv$とおくと

$$u + iv = z^3 = (x + iy)^3 = x^3 - 3xy^2 + i(3x^2 y - y^3)$$

より $u = x^3 - 3xy^2, v = 3x^2 y -y^3$.

(b) Let $z = x + iy$, $w = u + iv$. Then

$$u + iv = \frac{z}{z+1} = \frac{z(\bar{z} + 1)}{\vert z + 1\vert^2} = \frac{\vert z\vert^2 + z}{\vert z+1\vert^2}$$

Note that $\vert z\vert^2 = x^2 + y^2$.

$$u + iv = \frac{x^2 + y^2 + x + iy}{(x+1)^2 + y^2}$$

Therefore, $u = \frac{x^2 + y^2 + x}{(x+1)^2 + y^2}, v = \frac{y}{(x+1)^2 + y^2}$.

(c) Let $z = x + iy$, $w = u + iv$. Then

$$u + iv = \frac{z - i}{z + i} = \frac{(z-i)(z-i)}{\vert z + i\vert^2} = \frac{(x + i(y-1))^2}{\vert z+i\vert^2}$$

Note that $\vert z+i\vert^2 = x^2 + (y+1)^2$. Then

$$u + iv = \frac{x^2 - (y-1)^2 + 2ix(y-1)}{x^2 + (y+1)^2}$$

Therefore, $u = \frac{x^2 - (y-1)^2}{x^2 + (y+1)^2}, v = \frac{2x(y-1)}{x^2 + (y+1)^2}$.