Question

Medal to the first person to answer correctly! Evaluate this:

Answer 1

\[\LARGE \Re({ e^z})\]

Answer 2

whats r and whats z

Answer 3

R means evaluate the real part, z is a complex number.

Answer 4

Whoops I messed up, oh well. I meant to write \[\LARGE \Re({ e^{iz}})\] But it's too late now lol

Answer 5

so R(a+bi)=a?
that is the definition of that function?

Answer 6

I tried this the other day and I got \[
e^{-\Im(z)} \cos(\Re(z))
\]

Answer 7

But I don't think it was right.

Answer 8

where a and b are real

Answer 9

Not quite, the real and imaginary parts exist separately as a linear combination as a+bi. a is the real and b is the imaginary.

Answer 10

what does evaluates the real part mean

Answer 11

So if I ask for the real part of z^2 then I would do this \[\Large z^2 = (a+bi)^2 = a^2+b^2 -i2ab\] So we write \[\Large \Im(z^2) = -2ab \\ \Large \Re(z^2) = a^2+b^2\]

Answer 12

so R(a+bi)=a then?

Answer 13

Yep

Answer 14

ok i got confused for a second when you said not quite

Answer 15

Oh I was answering wio, sorry for the confusion

Answer 16

\[e^{i (a+bi)}= \cos(a+bi)+i \sin(a+bi)\]
so then we would use sum identities I suppose
this might turn ugly

Answer 17

Okay, now I get: \[
\cos(\Re(z))\cosh(\Im(z))-\cos(\Re(z))\sinh(\Im(z))
\]

Answer 18

\(e^{-\mathcal{I}(z)}\cdot \cos(\mathcal{R}(z))\)

Answer 19

Nope to all. =P There is only one real and imaginary part and they exist like this and this only \[\Large f(z) = \Re(z) + i \Im(z)\]

Answer 20

\[e^{i(a+bi)}=\cos(a)\cos(bi)-\sin(a)\sin(bi)+i(\sin(a)\cos(bi)+\cos(a)\sin(bi)) \\ =\cos(a)\cosh(b)-\sin(a) \cdot i \sin(x)+i(\sin(a)\cosh(b)+\cos(a) \cdot i(\sin(b)) \\ =\cos(a)\cosh(b)-\sin(b)\cos(a)-\sin(a)\sin(x)i+\sin(A)\cosh(b)i\]

Answer 21

@freckles is on the right path, but he's forgotten exponent rules. =P

Answer 22

so the real part is cos(a)cosh(b)-sin(b)cos(a)?

Answer 23

if i didn't make a mistake

Answer 24

oh

Answer 25

freckles is she

Answer 26

freckles can be a he
who knows

Answer 27

and yes I totally forgot the exponent thingy

Answer 28

This equation is not true\[\Large f(z) = \Re(z) + i \Im(z)\]

Answer 29

@freckles ,whatever this unicorn may be, you might like that you can derive trig identities this way!\[\Large e^{i(\theta + \phi)} = \cos(\theta + \phi) +i \sin(\theta + \phi) \\ \Large e^{i \theta}e^{i \phi}=(\cos \theta +i \sin \theta) (\cos \phi + i \sin \phi)\]

Answer 30

@wio lol unfortunately you're wrong.

Answer 31

Every complex function has a real and imaginary part.

Answer 32

Let\(f(z) = z^2\). Then \(f(i) = -1\), and not \(\Re(i) + \Im(i)i = (0)+(1)i = i\).

Answer 33

I'm fairly certain that: \[
\Re(e^z)=\cos(\Re(z))\cosh(\Im(z))-\cos(\Re(z))\sinh(\Im(z))
\]

Answer 34

One way to find an explicit formula for Re and Im is to just realize that the conjugate negates the imaginary part. \[\Large \bar z = a-bi\]\[\Large z + \bar z = 2a \\...so \\ \Large \Re(z) = \frac{z+ \bar z}{2}\] Similarly we have \[\Large \Im (z) = \frac{z - \bar z}{i2}\]

Answer 35

\[e^{i(a+bi)}=e^{ai+bi^2}=e^{-b}e^{ai}=e^{-b}(\cos(a)+i \sin(a)) \\ e^{-b} \cos(a)+i e^{-b} \sin(a)\]
this seems to easy
did i fail

Answer 36

the real part is e^(-b)cos(a)

Answer 37

Yes perfect!

Answer 38

which is what I wrote above

Answer 39

And I wrote it before watchmath wrote it, but it is wrong, lol

Answer 40

i don't know what the fancy stuff is :p

Answer 41

I was asking for Re(z), if you give me an answer in terms of R(z) then it isn't the answer.

Answer 42

No, you were asking for \(\Re(f(z))\) and you are probably going to need \(\Re(z)\) and \(\Im(z)\).

Answer 43

what do you mean? it is the same a = imaginary part of z= I(z) and b=real part of z =R(z)
so \(e^{-I(z)}\cos(Re(z))\) is the same

Answer 44

Remember that \(a=\Re(z)\) and \(b=\Im(z)\)

Answer 45

Yeah you're right, I guess I'll just have to give you all a third of a medal.

Answer 46

However since 1/3 rounds down to 0, I guess nobody gets medals. =(

Answer 47

But @Kainui , I'm not sure that @freckles and @watchmath .

In fact, I think this is a part where Euler's equation somehow breaks down.

Answer 48

Okay, actually, let me explain a bit better

Answer 49

What @freckles gave was:\[
f(z) = e^{iz}
\]

Answer 50

Oh? Keep going.

Answer 51

There are two ways to go about doing this problem, one is use exponent algebra, and the other is to use trig sum formula and hyperbolic trig functions

Answer 52

They should lead to the same result, but I want to try it out on both to be sure.

Answer 53

ok, I just gave wio a medal :D

Answer 54

Oh wio by the way I just discovered the derivation of the Cauchy-Riemann equations, it's almost obvious after you see it lol.

Answer 55

Here is one way to do it:\[
e^{z} = e^{i((a+ib)/i)} = e^{i(b-ai)} = e^{bi+a}=e^{a}e^{ib} = e^a\cos(b)+e^ai\sin(b)
\]

Answer 56

Wait, wtf did I just do, that was sort of a waste.

Answer 57

Just take the partial derivative\[\Large \frac{\partial f}{\partial x}=\frac{\partial f}{\partial z}\frac{\partial z}{\partial x} = \frac{\partial f}{\partial z} \\ \Large \frac{\partial f}{\partial y}=\frac{\partial f}{\partial z}\frac{\partial z}{\partial y} = \frac{\partial f}{\partial z} i\]

Then set the right hand sides equal after dividing the second one by i to get: \[\Large \frac{ \partial f}{\partial x} = \frac{1}{i} \frac{\partial f}{\partial y}\] and there are your two equations after you plug in f=u+iv.

Answer 58

Lol, wtf? Okay, but then:\[
e^z = e^{i(z/i)} = \cos(z/i)+i\sin(z/i)
\]

Answer 59

I think it is kind of nice, then we get relations like this: \[\Large \cos(ix)=\cosh(x)\] So you can think of how a circle is perpendicular to a hyperbola in a conic section maybe to make sense of this?

Answer 60

\[
\cos(b-ai)+i\sin(b-ai) = \cos(b)\cosh(a)+\ldots
\]Don't you think it is weird how we didn't have \(\cosh\) in our previous solution?

Answer 61

@Kainui There is something fishy going on, don't you think?

Answer 62

Do you agree with this?\[
\Re(e^{a+bi}) = e^a\cos(b)
\]

Answer 63

It certainly makes sense.

Answer 64

No, that's what I just said right before that. Just think of it as being on conic sections.

Answer 65

Wait, so they're not equal?

Answer 66

Huh, no they are equal. \[\Large \cos(ix)=\cosh(x)\]

Answer 67

Okay, well hold on, I want to work this out.

Answer 68

First, it is tempting to say: \[
e^z=e^{a+bi} = e^a\bigg(\cos(b)+i\sin(b)\bigg)
\]But consider: \[
e^{z} = e^{i(z/i)} = \bigg(\cos(z/i)+i\sin(z/i)\bigg) = \bigg(\cos(b-ai)+i\sin(b-ai)\bigg)
\]Will they result in same answer? Maybe so.

Answer 69

I don't know I am just making this up as I go, but it seems like we can figure it out from this picture: