Derivative of a Complex Function

[Drakkith]

Homework Statement

Find the derivative of $f(z)=\frac{1.5z+3i}{7.5iz-15}$

Homework Equations

The Attempt at a Solution

I had no difficulty using the standard derivative formulas to find the derivative of this function, but the actual result, that the derivative is zero, is confusing. For real functions, the derivative is usually only zero if your function is just a number with no variables. However, here we obviously have variables in the form of $x$ & $y$ in the two z's, but somehow the derivative is still zero.

Taking the partial derivatives gives me:
$\frac{\partial z}{\partial x} = \frac{-11.25y-22.5-22.5iy}{-56.25x^2-112.5ixy+56.25y^2-225i+225}$

$\frac{\partial z}{\partial y} = \frac{33.75ix+11.25x-67.5}{-56.25x^2-112.5ixy+56.25y^2-225i+225}$

If my math is correct, does this not show that the derivative changes and isn't zero?

What am I missing?

 

[Ray Vickson]

Homework Statement

Find the derivative of $f(z)=\frac{1.5z+3i}{7.5iz-15}$

Homework Equations

The Attempt at a Solution

I had no difficulty using the standard derivative formulas to find the derivative of this function, but the actual result, that the derivative is zero, is confusing. For real functions, the derivative is usually only zero if your function is just a number with no variables. However, here we obviously have variables in the form of $x$ & $y$ in the two z's, but somehow the derivative is still zero.

Taking the partial derivatives gives me:
$\frac{\partial z}{\partial x} = \frac{-11.25y-22.5-22.5iy}{-56.25x^2-112.5ixy+56.25y^2-225i+225}$

$\frac{\partial z}{\partial y} = \frac{33.75ix+11.25x-67.5}{-56.25x^2-112.5ixy+56.25y^2-225i+225}$

If my math is correct, does this not show that the derivative changes and isn't zero?

What am I missing?

You are missing the fact that $f(z)$ is equal to a complex constant $c$ for all $z = x+iy$. If you look a bit harder it becomes clear by inspection.

 

[Drakkith]

You are missing the fact that $f(z)$ is equal to a complex constant $c$ for all $z = x+iy$. If you look a bit harder it becomes clear by inspection.

Hi Ray. Would you mind elaborating? I'm afraid I don't see anything at the moment.

 

[fresh_42]

Hi Ray. Would you mind elaborating? I'm afraid I don't see anything at the moment.

What happens if you multiply the nominator by $5i\,$? Thus you can write $f(z)=\dfrac{1}{5i} \cdot \dfrac{\ldots}{\ldots}$.

 

[Drakkith]

What happens if you multiply the nominator by $5i\,$? Thus you can write $f(z)=\dfrac{1}{5i} \cdot \dfrac{\ldots}{\ldots}$.

I see. Thanks guys.