The Cauchy Riemann Equations for this Function

[RJLiberator]

Homework Statement

Verify that each of the following functions is entire:
f(z)=(z^2-2)e^(-x)e^(-iy)

Homework Equations

The Cauchy Riemann equations u(x,y) = ______ and v(x,y) = ______
u_y=-v_x
u_x=v_y

The Attempt at a Solution

So, I've done a few of these problems and understand that to prove it's entire we need to show that the C-R equations exist for all z and are satisfied. Also, noting that Ux Uy Vx Vy are continuous.

My problem is, I'm not sure how to get this into correct form. What is u(x,y) =?
Do I distribute first? How to change this function into something that works?

Would it be
(z^2-2)/(e^x*e^iy)
Is that a good start? :/

 

[Mark44]

Homework Statement

Verify that each of the following functions is entire:
f(z)=(z^2-2)e^(-x)e^(-iy)

Homework Equations

The Cauchy Riemann equations u(x,y) = ______ and v(x,y) = ______
u_y=-v_x
u_x=v_y

The Attempt at a Solution

So, I've done a few of these problems and understand that to prove it's entire we need to show that the C-R equations exist for all z and are satisfied. Also, noting that Ux Uy Vx Vy are continuous.

My problem is, I'm not sure how to get this into correct form. What is u(x,y) =?
Do I distribute first? How to change this function into something that works?

Would it be
(z^2-2)/(e^x*e^iy)
Is that a good start? :/

I don't believe so.

For starters, write z = x + iy, and split $(z^2 - 2)e^{-x}e^{-iy}$ into its real and imaginary parts. u(x, y) is the real part of f(z) and v(x, y) is the imaginary part.

A useful formula that you omitted, and that might come in handy, is $e^{it} = \cos(t) + i\sin(t)$.

 

[RJLiberator]

Thanks Mark.

So we do have to expand the z then.

So this would mean:

(x^2+2yxi-y^2-2)/(e^xcos(y)+e^x(isin(y)))

The thing that is throwing me off here is the negative exponents, I'm not sure what to do with those, so I put the term in the denominator.

That doesn't seem right however.

 

[Mark44]

Moving them to the denominator is no help. Leave the e^-x factor as is, and rewrite the e^-iy factor as below.
$e^{-x}e^{-iy} = e^{-x}(\cos(y) - i\sin(y))$

 

[HallsofIvy]

Rather than put "$e^xcos(y)+e^x(isin(y))$" in the denominator, I think I would write $e^{-x- iy}= e^{-x}e^{-iy}= e^{-x}(cos(-y)+ i sin(-y))= e^{-x}(cos(x)- i sin(x))$

 

[RJLiberator]

That was indeed the element of this question that I was missing out/overlooked.

So with that in play, we see that
u(x,y) = (x^2-y^2-2)e^(-x)cos(x)+2ye^(-x)sin(y)
v(x,y) = 2ye^(-x)cos(x)-(x^2-y^2-2)e^(-x)sin(y)

 

[RJLiberator]

Re: this thread, I found out that the composition of two functions that are entire is entire.

So One can look at (z^2-2) and e^(-x)e^(-iy) as two different functions, evaluate their cauchy-riemann equations and see that each is entire to conclude the composition is entire.

=)

 

[WWGD]

But notice you are not composing the functions, you are multiplying them ( though the result is true that the product of entires is entire )