Cauchy-Riemann condition

• chwala
In summary, the conversation discusses verifying whether the function f(z)= z^3-5iz+√7 satisfies the Cauchy Riemann equations. The attempt at a solution includes substituting for z=x+iy and showing that both sides of the Cauchy Riemann partial differential equations hold, but the person is seeking alternative methods of solving. The expert suggests using general principles, such as the fact that sums and products of functions satisfying CR also satisfy CR, to avoid messy calculations. The person also mentions the harmonic condition as a necessary but not sufficient condition for satisfying CR.

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.

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Delta2

chwala said:

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.

chwala
chwala said:
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
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.

chwala
Mark44 said:
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. that's a summary of my working, or you want to see the full working?

FactChecker said:
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##

jambaugh said:
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,

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.]

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chwala
jambaugh said:
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.

chwala said:
thats exactly what i did. i substituted for ##z=x+iy##
What you did is not the same as applying general principles. You don't have to do the messy calculations if the principles have been proven. You have to have proven the principle that sums and multiplications of functions satisfying CR will also satisfy CR. Then just state that your function is a combination of sums and multiplications of simple functions satisfying CR and apply the principle.

chwala
chwala said:
why? i have shown that both sides of the cauchy riemann pde hold. that's a summary of my working, or you want to see the full working?
kindly note pde means partial differential equations...just for clarity