Algebraic form for a function of a complex variable.

Let $f$ be a function of a complex variable, defined over a domain $\mathcal{D}$ in $\mathbb{C}$ . We write $z=x+iy$ , where $x$ and $y$ are real numbers, i.e. $z$ is written in algebraic form. We can write $f(z)$ in algebraic form too, i.e.

$$f(z)= \underbrace{u(x,y)}_{=\text{Re}(f(z)} +i \underbrace{v(x,y)}_{=\text{Im}(f(z)}$$ (2.1)

Example 2.2.1   Let $f(z)=z^2$ , for $z \in \mathbb{C}$ . With $z=x+iy$ , where $x$ and $y$ , as above, we have:

$$f(z)=(x+iy)^2 = \underbrace{x^2-y^2}_{=u(x,y)} +i \; \underbrace{2xy}_{=v(x,y)}$$

Example 2.2.2   Let $f(z)=\frac 1z$ , for $z \in \mathbb{C} - \{ 0 \}$ . With $z=x+iy$ , where $x$ and $y$ , as above, we have:

$$f(z)=\frac 1z = \frac {1}{x+iy} = \frac {x-iy}{(x+iy)(x-iy)} = \frac {x-iy}{x^2+y^2}$$

Thus

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

Conversely, if we have a function given in algebraic form (v.s. 3)

$$f(z)= \underbrace{u(x,y)}_{=\text{Re}(f(z)} +i \underbrace{v(x,y)}_{=\text{Im}(f(z)}$$

we can compute a ``closed'' form for $z$ , using the following remark:

$$\begin{cases}z=x+iy \ \overline{z}=x-iy \end{cases} \Longrig... ...\overline{z}) \ y= \frac {1}{2i} (z- \overline{z}) \end{cases}$$ (2.2)

Example 2.2.3   Let $f(z)=x^2+iy^2$ . Using Euler formulas, i.e. Equation (4), we have:

$$f(z) = \left( \frac 12 (z+ \overline{z}) \right) ^2 + i \left( \frac {1}{2i} (z- \overline{z}) \right) ^2$$

$$\quad = \frac 14 (z^2+2z \; \overline{z} + \overline{z}^2 ) - i \; \frac 14 ( z^2 - 2z \; \overline{z} + \overline{z}^2 )$$

$$\quad = \frac 14 (1+i)(z^2+\overline{z}^2) + \frac 12 ( 1-i)z \; \overline{z}.$$

Of course, the converse process is possible, i.e. for a function given by a formula like $f(z)=u(x,y)+i\; v(x,y)$ , Euler formulas can be used to give an expression in $z$ and $\overline{z}$ for $f(z)$ .

Example 2.2.4   Let $f(z)=x^2+ i\; y^2$ . We use Euler formulas:

$$\begin{cases}x=\frac 12 (z+ \overline{z})\ y=\frac{1}{2i} (z... ...2=-\frac 14 (z^2-\overline{z}^2-2z\; \overline{z}) \end{cases}.$$

Thus,

$$f(z) =\frac 14 \left[ z^2-\overline{z}^2+2z\; \overline{z}) -i \; \left( (z^2-\overline{z}^2-2z\; \overline{z}) \right) \right]$$

$$\quad = \frac 14 \left[ z^2-\overline{z}^2+2\vert z\vert^2) -i \; \left( (z^2-\overline{z}^2-2\vert z\vert^2) \right) \right].$$