complex analysis

[meraman88]

hi, can you plz help me with complex analysis problem:

The function f(z) = e^(z+i*pi) has infinitely many points in the fiber of each point in its range.
(A) Find four points that map to 1
(B) The natural inverse of f(z), say g(z) maps each point in its domain to infinitely points, as with log(z). For example, g(1) is infinitely many points, including the four you provided in the first part of the problem.
i. Choose a branch of g(z)
ii. Sketch the domain of this branch
iii. Sketch the range of this branch
IV, Indicate which point in the set g(1) is the image of 1 under your branch.

thanks in advance who help me..

[meraman88]

plz someone help with this

[AlphaNumeric]

This looks like homework. What are your attempts so far?

[meraman88]

i have tried using euler's identity formula: cos+ i*sin
but not understanding how to start with it. i am starter so still not understanding concept correctly.
so plz help if you can understand it. i will be thankful to you.

[AlphaNumeric]

Use Euler's formula and then equate the real and imaginary parts on each side. Then use what you know about sin and cos.

[meraman88]

can you plz show me solution of one so that by understanding that i can solve remaining part of my problem?

[prometheus]

I particularly like that your dedication to getting your homework done extends to posting on peoples visitor messages.

[meraman88]

thanks for atleast commenting but bro i dont think asking help is a bad thing, i am asking solution for my problem, rather asking to guide me so that i can complete it.

by the way, have a nice day to you mr. prometneus.

[AlphaNumeric]

I particularly like that your dedication to getting your homework done extends to posting on peoples visitor messages.

I got a private message.

can you plz show me solution of one so that by understanding that i can solve remaining part of my problem?

So what is your solution to the first part? Let's get that right first.

I find it a little odd you're doing complex analysis and branch cuts but you struggle to solve . You really shouldn't be doing branch cuts if you don't know how to apply Euler's formula to that.

[meraman88]

i have said earlier sir that i have just started this course so it's normal that i am finding it hard right now.
so i am asking for guidance, not whole solution. if you can do it, i will be gr8 for me for understanding this

[prometheus]

Ok. AN suggested you use Euler's formula to rewrite the equation (I assume ):



So how can you simplify this and what do you know about the cos and sin function?

[AlphaNumeric]

A useful thing to note when trying to work out the value of is that and for all real .

[meraman88]

hi, thanks for help to all.

i have tried to solve this way, can you tell is it right or wrong? is the sequence and presentation is right? and if possible solution for next step.

Ans. (a).
Here, function f(z) = e^(z+i*pi),
First write down given function like below to simplify it,
f(z) = e^(z+i*pi) = e^(i*pi)e^z
this will become,
e^(i*pi)e^z = -e^z
And as we have e^(n*2*pi*i)=1, for any integer n,
So,
f(i*pi)=1, f(2*i*pi)=1, f(3*i*pi)=1, f(4*i*pi)=1, etc.
Ans. (b).
Now, to find the natural inverse of f(z)=e^z, we need to solve the equation e^z=w, given w .
The answer is z=log(w).
If w=r e^(i*Ө), then z= log(r)+i* Ө = log(|w|)+I*Arg(w).
This is many value as Arg is many valued.
The natural inverse of f(z) is therefore:
(z+i*pi)=log(w)
So , z=log(w) – i*pi, hence, g(z) = log(w) – i*pi


thanks