# Cauchy.Riemann equations

1. Jun 2, 2014

### MissP.25_5

Hello.
Here I attached the question and what I have done so far. I have used the cauchy.riemann equations to check if the functions are analytical or not. The first one turns out to be analytical, but I don't know how to find its derivative function and how do I express it as a z function?

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2. Jun 2, 2014

### CAF123

Hint: Rewrite the first equation as $$f(x,y) = e^x x (\cos y + i \sin y) + e^x y (i\cos y - \sin y)$$

Now do some further manipulations and simplify, so as to be left with something purely in terms of $z = x+iy$.

3. Jun 2, 2014

### MissP.25_5

But what about this formula? I got this from a textbook. f'(z0) here is the derive function.

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4. Jun 2, 2014

### CAF123

Once you rewrite $f(x,y)$ as defined in the last post entirely in terms of $z$, then computing df/dz is arguably easier.

Use Euler's formula when simplifying $f(x,y)$.

Edit: I just saw your attachment. That is the definition of the derivative. But generally speaking there are techniques you should use such as product rule which are easier to deal with for more complicated functions.

5. Jun 2, 2014

### MissP.25_5

So the one I just attached is not a formula but actually just a definition of the derivative?

6. Jun 2, 2014

### CAF123

It can be used in practice too, but even with the most simplest of functions you'd better off using the power rule, product rule etc..

E.g assuming you are familiar with the real version, let $f(x) = x^2$. By power rule, $df/dx = 2x$. Done.

By the definition of the derivative, $$f'(x) = \lim_{h \rightarrow 0}\frac{(x+h)^2 - x^2}{h} = \lim_{h \rightarrow 0} \frac{2hx + h^2}{h} = 2x.$$
Then try it with $f(x) = \sqrt{x}$ and already more manipulations needed.

7. Jun 2, 2014

### MissP.25_5

Is this what you mean? Did I do it correctly?

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8. Jun 2, 2014

### CAF123

Close, but $-1/i = i$, so you messed up a sign. Other than that, looks fine. Now use the fact that $z = x+iy$ to obtain a function all in terms of z.

Note also by simply factoring out $i$ from the second term you get $ie^x y(\cos y + i\sin y) = ie^x ye^{iy}$, so no need to mess about with more definitions.

9. Jun 2, 2014

### MissP.25_5

Now is this correct?
So, next how do I obtain a function?

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10. Jun 2, 2014

### CAF123

Yes, it is correct. But you can simplify further. $e^a e^b = \dots$ and use that z = x+iy.

11. Jun 2, 2014

### MissP.25_5

You mean like this? Is that the final answer?

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• ###### IMG_6407.jpg
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12. Jun 2, 2014

### CAF123

One last simplification you can make. Make sure you have only z in your expression (no appearances of x or y). Then are you after df/dz?

13. Jun 2, 2014

### MissP.25_5

Aha! I think I got it. Is this right?

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14. Jun 2, 2014

### CAF123

Yeah, it is good now.