# Homework Help: Expressing f(z) in terms of z

1. Jul 13, 2014

### MissP.25_5

Hello! I am stuck at the final step. How do I get x+iy from the equation? Help!

I am so sorry for posting this question in a picture instead of writing it out, because I don't know how to write equations on here.

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2. Jul 13, 2014

### Staff: Mentor

I see a small mistake: on the 4th line, shouldn't you be taking out the common factor 1/4?

Check your exponential equivalents for sine, cosine, etc., on line 3. There's your mistake.

Last edited: Jul 13, 2014
3. Jul 13, 2014

### MissP.25_5

why 1/4 ?

4. Jul 13, 2014

### Mentallic

You substituted incorrectly.

$$\cos{z}=\frac{e^{iz}+e^{-iz}}{2}$$

$$\sin{z}=\frac{e^{iz}-e^{-iz}}{2i}$$

And you should factor out a value on the 4th line because you want your expression to be of the form

$$k\left(\frac{e^{iz}-e^{-iz}}{2i}\right)$$
(or some similar form in the brackets)
where k is some complex number, and z is a complex number. Your expression would then simplify to $k\sin{z}$

5. Jul 13, 2014

### MissP.25_5

But z=x+iy. How do I change i(x^2-y^2) into z? I can't figure out how to rearrange it.

6. Jul 13, 2014

### Mentallic

Well of course you should recognize that

$$z^2=(x+iy)^2=x^2-y^2+i 2xy$$

and what you have is VERY similar to this form. Since you made a mistake early on that I pointed out, your final result isn't going to work with the method I'm hinting at here, but when you fix that up then it'll fall into place.

$$i(x^2-y^2)-2xy$$

Should be quite easily converted into a function of z by observing the z2 result.

7. Jul 14, 2014

### MissP.25_5

Ok, so I have corrected my mistakes but I got stuck here. How do I continue?

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8. Jul 14, 2014

### Mentallic

Now expand like you did earlier and simplify where possible.

9. Jul 14, 2014

### MissP.25_5

I did that but still couldn't get the answer. Hold on, I will show you my work later.

10. Jul 14, 2014

### Mentallic

Did you end up with

$$-i\left(\frac{e^{w}-e^{-w}}{2}\right)$$

where

$$w=i(x^2-y^2)-2xy$$

??

11. Jul 14, 2014

### MissP.25_5

I don't know what to do next. How do I simplify this further?

Last edited: Jul 14, 2014
12. Jul 14, 2014

### MissP.25_5

Sorry, did a mistake earlier. Here's what I got.

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13. Jul 14, 2014

### Mentallic

Sorry, I only just glossed over your previous upload and didn't spot the error.

$$\sinh{z}=\frac{e^z-e^{-z}}{2}$$

while you had an i in the denominator.

14. Jul 14, 2014

### MissP.25_5

Oh yes, that was my mistake. But then what happens to the i outside of the bracket?

15. Jul 14, 2014

### Mentallic

The first 4 terms have a constant factor of

$$\frac{1}{4i}$$

while the right 4 are

$$\frac{i}{4}$$

How do these two numbers relate to each other?

16. Jul 14, 2014

### MissP.25_5

They're minus to each other. Sorry, I don't know how to say that mathematically.

17. Jul 14, 2014

### Mentallic

Right!

$$\frac{1}{4i}=\frac{i}{4(-1)}=\frac{-i}{4}$$

and so now you can continue to simplify the expression!

18. Jul 14, 2014

### MissP.25_5

Simplified it, then what? I am still stuck here.

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19. Jul 14, 2014

### Mentallic

Check post #10.

20. Jul 14, 2014

### MissP.25_5

Uhmm, how do I change w into z? Can I just simpy do this?

exp(i*(ia-b)) ?? Is it possible to multiply i just like that?

21. Jul 14, 2014

### Mentallic

I'm using w just as a place-holder. Just like a and b are place-holders.

Let w=a-b.

22. Jul 14, 2014

### MissP.25_5

Ok, but how do we get sin(a+ib) from it? I can't seem to rearrange it.

23. Jul 14, 2014

### Mentallic

Notice that -ia+b = -(ia-b)

so if we let w=-ia+b (or we could have also done w=ia-b) then

$$\frac{i}{2}(e^{-ia+b}-e^{ia-b})$$
$$=\frac{i}{2}(e^{-ia+b}-e^{-(-ia+b)})$$
$$=\frac{i}{2}(e^{w}-e^{-w})$$
$$=i\left(\frac{e^{w}-e^{-w}}{2}\right)$$
$$=i\cdot\sinh(w)$$

Let's substitute back now since we don't want w.

$$w=-ia+b$$
$$=-i(x^2-y^2)+2xy$$
$$=-i\left((x^2-y^2)+i2xy\right)$$

And what is $x^2-y^2+i2xy$ equal to in terms of z?

24. Jul 14, 2014

### MissP.25_5

z2! The answer is sin(z2), right? Thank you for your explanations! By the way, could you tell me how you write the equations like that? What format do you use?

25. Jul 14, 2014

### Mentallic

$$w=-iz^2$$

and so

$$f(z)=i\cdot\sinh{(-iz^2)}$$

but

$$\sin(a)=\frac{\sinh(ia)}{i}$$

and so

$$f(z)=-\frac{\sinh(i(-z^2))}{i}$$

$$=-\sin(-z^2)=\sin(z^2)$$

so yes, you were correct. You could have instead chosen w such that we end up with sin() rather than sinh() and avoided all of this simplification though.

I use tex tags, check out
https://www.physicsforums.com/showpost.php?p=3977517&postcount=3