# Cauchy-Riemann condition

Gold Member

## Homework Statement

Verify $f(z)= z^3-5iz+√7$ satisfies cauchy riemann equations.

## The Attempt at a Solution

seeking alternative method
$f(z)= (x^3+5y-3xy^2+√7)+ (3x^2y-y^3-5x)i$
$∂u/∂x = 3x^2-3y^2 = ∂v/∂y$
$∂v/∂x=6xy-5= -∂u/dy$
hence satisfies.

Last edited by a moderator:

Related Calculus and Beyond Homework Help News on Phys.org
jambaugh
Gold Member

Mark44
Mentor

## Homework Statement

Verify $f(z)= z^3-5iz+√7$ satisfies cauchy riemann equations.

## The Attempt at a Solution

seeking alternative method
Why? Is there some reason that the work below isn't satisfactory?
chwala said:
$f(z)= (x^3+5y-3xy^2+√7)+ (3x^2y-y^3-5x)i$
$∂u/∂x = 3x^2-3y^2 = ∂v/∂y$
$∂v/∂x=6xy-5= -∂u/dy$
hence satisfies.

Dick
Homework Helper
seeking alternative method
If you think about expressing $f=u+iv$ in terms of $z$ and $\bar{z}$ instead of $x$ and $y$, it's easy to show that the CR equations say that $\frac{\partial{f}}{\partial{\bar{z}}}=0$. So any function that depends only on $z$ will satisfy CR.

FactChecker
Gold Member
There is a simple way to see it immediately if you know that the sums and products of functions that satisfy CR also satisfy CR. Constants and f(z)=z satisfy CR. Your equation is a combination of sums and products satisfying CR, so it also satisfies CR.

Gold Member
Why? Is there some reason that the work below isn't satisfactory?
why? i have shown that both sides of the cauchy riemann pde hold. thats a summary of my working, or you want to see the full working?

Gold Member
There is a simple way to see it immediately if you know that the sums and products of functions that satisfy CR also satisfy CR. Constants and f(z)=z satisfy CR. Your equation is a combination of sums and products satisfying CR, so it also satisfies CR.
thats exactly what i did. i substituted for $z=x+iy$

Gold Member
My method is correct, its not an attempt. I was proving or rather to show that the given $f(z)$ satisfies the Cauchy Riemann equations. I am simply asking for alternative ways of solving. regards,

jambaugh
Gold Member
Well as the Cauchy Riemann conditions are directly verifiable there's not much else you can do but directly verify them. There are consequential conditions like the harmonic condition:
$$\Delta u = \Delta v = 0$$
but while this is a necessary condition it is not sufficient. It may be satisfied for functions failing to satisfy CR.
[edit: oops for the typo... now corrected.]

Last edited:
Gold Member
Well as the Cauchy Riemann conditions are directly verifiable there's not much else you can do but directly verify them. There are consequential conditions like the harmonic condition:
$$\Delt u = \Delta v = 0$$
but while this is a necessary condition it is not sufficient. It may be satisfied for functions failing to satisfy CR.
.....directly verifiable.........thank you.

FactChecker
thats exactly what i did. i substituted for $z=x+iy$