Calculus in the Complex Plane

COMP116 Tutorials 2021-03-15

Expressing Complex Functions in Terms of $u$ and $v$.

Calculate for $z=x+iy$

1. $f(z)=z^3$

   $(x+iy)(x+iy)(x+iy)$

   $(x^2+2ixy-y^2)(x+iy)$

   $x^3+x^2iy+2ix^2y-2xy^2-xy^2-iy^3$

   $u(x,y)=x^3-3xy^2$

   $v(x,y)=3x^2y-y^3$

2. $f(z)=\vert z\vert+ z^2$

   $(x^2+y^2)+(x+iy)^2$

   $(x^2+y^2)+x^2+2ixy-y^2$

   $2x^2+2ixy$

   $u(x,y)=2x^2$

   $v(x,y)=2xy$

Cauchy-Riemann Conditions

1. $f(z)=z^3$

   $u(x,y)=x^3-3xy^2$

   $v(x,y)=3x^2y-y^3$

   $\frac{\partial u}{\partial x}=3x^2-3y^2$

   $\frac{\partial v}{\partial y}=3x^2-3y^2$

   This part is valid.

   $\frac{\partial u}{\partial y}=-6xy$

   $\frac{\partial v}{\partial x}=6xy$

   This part is also valid.

2. $f(z)=\vert z\vert+ z^2$

   $u(x,y)=2x^2$

   $v(x,y)=2xy$

   $\frac{\partial u}{\partial x}=4x$

   $\frac{\partial v}{\partial y}=2x$

   The first condition is not met.

   This means that this function is only differentiate-able at $x=0$.

   $\frac{\partial u}{\partial y}=0$

   $\frac{\partial v}{\partial x}=2y$

   This condition is also not met.

First Derivative

The first derivative of $f(z)=z^3$ is:

\[f'(z)=3z^2\]