How can I express f(z) in terms of z?

[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.

 

[NascentOxygen]

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.

 

[MissP.25_5]

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

why 1/4 ?

 

[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}

 

[MissP.25_5]

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}

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

 

[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.

 

[MissP.25_5]

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.

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

 

[Mentallic]

Now expand like you did earlier and simplify where possible.

 

[MissP.25_5]

Now expand like you did earlier and simplify where possible.

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

 

[Mentallic]

Did you end up with

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

where

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

??

 

[MissP.25_5]

Now expand like you did earlier and simplify where possible.

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

 

[MissP.25_5]

Did you end up with

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

where

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

??

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

 

[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.

 

[MissP.25_5]

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.

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

 

[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?

 

[MissP.25_5]

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?

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

 

[Mentallic]

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

Right!

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

and so now you can continue to simplify the expression!

 

[MissP.25_5]

Right!

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

and so now you can continue to simplify the expression!

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

 

[Mentallic]

Check post #10.

 

[MissP.25_5]

Check post #10.

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?

 

[Mentallic]

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?

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

Let w=a-b.

 

[MissP.25_5]

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

Let w=a-b.

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

 

[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?

 

[MissP.25_5]

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?

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?

 

[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