Evaluate complex value and domain of analytic function

[yy205001]

Homework Statement

Can anyone help me to check whether the first two are correct or not?
For the third question, i may need some help on it.

Evaluate the following if they exist:
a) sin(3Logi)
b) (1+i)^(1/3-i) (principal value)

Determine the largest open set in which the function Log(1-z^n) is analytic. Here n is a positive integer.

Any help is appreciated!

Homework Equations

Log(z) = Log(abs(z)) + iArg(z)
z^a=exp(a*Log(z)) a, z\inℂ

The Attempt at a Solution

a) sin(3*Log(i))
= sin[3(Log(abs(i)) + iArg(i))]
= sin[3(Log(1)+i(pi/2))]
= sin[3(0+i*pi/2)]
= sin[i*3pi/2]
= -i*sinh[i*(i*3pi/2)]
= -i*sinh[-3pi/2]
= i*sinh[3pi/2]

b)(1+i)^(1/3-i) principal value
= exp[(1/3-i)*Log(1+i)]
= exp[(1/3-i)*(Log(abs(1+i))+iArg(1+i))]
= exp[(1/3-i)*(Log(sqrt(2))+i(pi/4))]

Third question:
Let z=x+iy.
First, I think i need to solve Re{1-z^n}≤0 and Im{1-z^n}=0 for x, y, then the largest domain is the ℂ except the set {z=x+iy| Re{1-z^n}≤0 and Im{1-z^n}=0}.

So,
→ 1-(x+iy)^n =0, then i can't anymore further. I tried to use binomial theorem on (x+iy)^n but just can't do it.

 

[HallsofIvy]

Homework Statement

Can anyone help me to check whether the first two are correct or not?
For the third question, i may need some help on it.

Evaluate the following if they exist:
a) sin(3Logi)
b) (1+i)^(1/3-i) (principal value)

Determine the largest open set in which the function Log(1-z^n) is analytic. Here n is a positive integer.

Any help is appreciated!

Homework Equations

Log(z) = Log(abs(z)) + iArg(z)
z^a=exp(a*Log(z)) a, z\inℂ

Okay, since you are asking for the principal value you don't need to do things like "plus a multiple of i\pi.

The Attempt at a Solution

a) sin(3*Log(i))
= sin[3(Log(abs(i)) + iArg(i))]
= sin[3(Log(1)+i(pi/2))]
= sin[3(0+i*pi/2)]
= sin[i*3pi/2]
= -i*sinh[i*(i*3pi/2)]
= -i*sinh[-3pi/2]
= i*sinh[3pi/2]

Yes, that looks correct. Personally, I would have used
sin(x)= \frac{e^{ix}- e^{-ix}}{2i}
but, of course, that is the same as your sinh(x).

b)(1+i)^(1/3-i) principal value
= exp[(1/3-i)*Log(1+i)]
= exp[(1/3-i)*(Log(abs(1+i))+iArg(1+i))]
= exp[(1/3-i)*(Log(sqrt(2))+i(pi/4))]

I think you can do a little better. first, of course Log(sqrt(2))= (1/2)Log(2).
Also you can multiply out in the exponential: {(1/6)Log(2)- (pi/4)}+ {(1/3)pi/4- (1/2)Log(2)}i.
And then use the fact that exp(a+ bi)= exp(a)(cos(b)+ isin(b)).

Third question:
Let z=x+iy.
First, I think i need to solve Re{1-z^n}≤0 and Im{1-z^n}=0 for x, y, then the largest domain is the ℂ except the set {z=x+iy| Re{1-z^n}≤0 and Im{1-z^n}=0}.

So,
→ 1-(x+iy)^n =0, then i can't anymore further. I tried to use binomial theorem on (x+iy)^n but just can't do it.

You are making a mistake in trying to deal with this in "Cartesian form". ln(z)= ln(re^{i\theta}= ln(r)+ i\theta, as you said before, is analytic as long as r> 0. That means this function is analytic as long as |1- z|> 0 which is exactly the same as saying z\ne 1.

 

[yy205001]

You are making a mistake in trying to deal with this in "Cartesian form". ln(z)= ln(re^{i\theta}= ln(r)+ i\theta, as you said before, is analytic as long as r> 0. That means this function is analytic as long as |1- z|> 0 which is exactly the same as saying z\ne 1.

HallsofIvy, so the answer is just z≠1??