Wow, the code did not turn out at all how I thought it was going to. I apologize for this confusion; I used the symbols button and just assumed they would translate to standard notation.
Homework Statement
Let C be a regular curve enclosing the distinct points w1,..., wn and let p(w)= (w-w1)(w-w2)...(w-wn). Suppose that f(w) is analytic in a region that includes C. Show that P(z)= (1/2\pii)∫(f(w)\divp(w))\times((p(w)-p(z)\div(w-z))\timesdw
is a polynomial of degree n-1...
Homework Statement
Give the partial fraction decomposition of 1/z4+z2
Homework Equations
The Attempt at a Solution
My question is about the final answer. The book gives the answer to be 1/z2+ 1/2i(z+i)- 1/2i(z-i). For my answer I keep getting a negative for both of the 1/2i...
Homework Statement
What I had to prove was sin(z1+z2)= sin(z1)cos(z2)+sin(z2)cos(z1). I did this simply using the e function definitions for sin and cos and it turned out fine. I then flipped to the back of my book to double check and they have a completely different method for proving it. The...
One last comment.. just because I realized the source of both our confusions as to what the other was doing. I was looking at the limit of integration wrong. Like utterly wrong lol. I thought it was pi divided by the quantity 2+i, when really, as you elucidated, it is (pi/2)+i. That is why I was...
Sorry, I was at a loss earlier and did that based on what a friend said because I had no idea. It did make sense to me though so it was my own error. In regards to the arithmetic: I was under the impression when you have i in the denominator, to put the fraction in a+bi form, you multiply the...
I did (pi/2+i)*(2-i)/(2-i)= (2pi-ipi)/5... specifically the 5 came from (2+i)(2-i)=(2^2)-i^2=5. I then took (2pi-ipi)/5 and plugged it in for z in the formula, which then has it being multiplied by 2i, and 2i*(2pi-ipi)/5= (4ipi+2pi)/5. Sorry for leaving all of that out.
I did (pi/2+i)*(2-i)/(2-i)= (2pi-ipi)/5... specifically the 5 came from (2+i)(2-i)=(2^2)-i^2=5. I then took (2pi-ipi)/5 and plugged it in for z in the formula, which then has it being multiplied by 2i, and 2i*(2pi-ipi)/5= (4ipi+2pi)/5. Sorry for leaving all of that out.
Alright I figured it was that formula. So (1/2)sin(2z) becomes (1/4i)(e^(2iz)-e^(-2iz)). Yes I do know how to do complex exponentials , but not very well at a practical level yet. I did take the i out of the denominator of 2+i (using the conjugate) and then after putting into the formula I have...
Homework Statement
Evaluate ∫cos(2z)dz from pi/2 to pi/2+i
Homework Equations
The Attempt at a Solution
I know the cos function is entire, thus independent of path and I just need to evaluate the end points. I also know when you integrate you get (1/2)sin2z. My only question is...
Ahh is it that I improperly multiplied the 4 through the equations? I think I should have multiplied through before I made the terms into squares, which would change for example 4y into 2y, etc.
Can you be more specific in regards to which part of my previous answer was wrong? Just so I know which part to look at. I see that I gave the ordered pair for the center backwards i.e (y,x) when it should be (x,y). I am not sure what you mean by handling the factor of 4 wrong though.
Alrighty. So now I have, after correcting the Re(Az) error and a little simplifying to get the 2's out of the denominators as well as producing the required 4B, (4y-2Im(A))2+ (4x+ 2Re(A))2= (ReA)2+ (ImA)2- 4B = \left|A^2\right|-4B. Since the LHS is greater then or equal to 0 the RHS must be the...