This is a textbook for an introductory course in complex analysis. It has been used for our undergraduate complex analysis course here at Georgia Tech and at a few other places that I know of.

Thank you for your advice. I'll study complex analysis some day.]]>

I was not assuming the result.

sinh(x)=exp(x)/2-exp(-x)/2 by defination

You're right.

George, I apologize. Sorry

In case you're interested, the reason why you need to exercise extreme caution when replacing any f(x) with f(z) is that, in the complex plane, you have to check for continuity on a *circle* centred on some f(z_0) rather than on the real line segment centred on some f(x_0).

Complex analysis throws up some truly amazing theorems, all based on the requirement that functions possess what might be called isotropic derivatives

]]>sinh(x)=exp(x)/2-exp(-x)/2 by defination

My software gives "Cosh[i pi/3]=1/2"

]]>By the way, i(x+iy)=-y+xi

I just post this to explain why sometimes Mathematica will give a result involving i when you assaign only an integration task, and that the coefficient of i is always very small. The reason, I guess, is that it did some circular to hypobolic shifting. And you can just ignore the imaginary part.

I admit the proof was insufficient.

]]>[ ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ ° ]

Copy and paste.

]]>George,Y wrote:

Sinh(iθ)= iSin(θ)Cosh(iθ)=Cos(θ)

for real x, cosh x = cos ix, and not as you wrote (how do you get thetas and stuff in here, without using LaTex?)

]]>I'm

astonished to see for far you guys haven't proved what I proposed!

Hence

sinh(iz)=[(1)-(2)]/2=isinz

cosh(iz)=[(1)+(2)]/2=cosz

Well you may be, and when I tell you why you won't like it!

First, you are assuming the result you are trying to prove (rarely a

good idea). You should really establish a relationship between the

circulars and the hyperbolics before taking your leap!

Second, even if we know that, if x is real, sinh ix = i sin x and cosh x = cos ix we need to do this.

Let z = x + iy, and replace x in the above by z. What do you get, for example for iz = i(x + iy)?

For cos iz?

For i sin z?

Hint: Use Osborne's Rule (know it?)

]]>Okay, this is how i got the two formulae.

Euler's formula( proven in #10)

substituding z with -z will get

simlified form:

Hence

sinh(iz)=[(1)-(2)]/2=isinz

cosh(iz)=[(1)+(2)]/2=cosz

]]>i wrote a terrain grapher in flash a while ago, when i made my complex number calculator, and i added a graphing function, and when i first tried it out, the first thing i noticed, was when graphing sinh and cosh, they were just 90 degree rotations in the complex plane

]]>Sorry ben. In an attempt to write a quick post, I was too brief.

No problem, math is fun, right?

Consider the equation:

But why should I? Where did it come from, other than the taylor series I showed?

And so it must be that:

Why? It could be, by your reasoning, that

Anyway, let's not fall out in public. Others seem to have lost interest anyway.

]]>Consider the equation:

We can show, as my last post did, that:

But we also know that:

So we can say:

And so it must be that:

]]>.

The 5 fundamental transcendentals, all in bed together (but no action!!)

]]>