Recent content by cragar

thanks for helping me clear it up.

I expanded the exponential as a Laurent series , then I multiplied by z^3, and then found the 1/z term , other ways won't work because its not a pole ,

so the exponential function, does not have a pole or a removable singularity, since it does not have a pole , I think I just have to expand it in a Laurent series to find the residue, I am trying to find the value of that integral over the circle |z|=5, and I am trying to use the residue theorem ,

Homework Statement use the residue theorem to find the value of the integral, integral of z^3e^{\frac{-1}{z^2}} over the contour |z|=5 The Attempt at a Solution When I first look at this I see we have a pole at z=0 , because we can't divide by zero in the exponential term. and a pole of...

ok just wanted to make sure

Homework Statement Use a laurent series to find the indicated residue f(z)=e^{\frac{-2}{z^2}} Homework EquationsThe Attempt at a Solution So I expand the series as follows 1-\frac{2}{z^2}+\frac{2}{z^4} ... my book says the residue is 0 , is this because there is no residue term ? the...

Homework Statement Determine the order of the poles for the given function. f(z)=\frac{1}{1+e^z} Homework EquationsThe Attempt at a Solution I know if you set the denominator equal to zero you get z=ln(-1) But if you expand the function as a geometric series , 1-e^{z}+e^{2z}... I...

oh right i mistyped the problem it has a z-3 factor in the original factor

Homework Statement expand f(z)=\frac{1}{z(z-1)} in a laurent series valid for the given annular domain. |z|> 3 Homework EquationsThe Attempt at a Solution first I do partial fractions to get \frac{-1}{3z} +\frac{1}{3(z-3)} then in the second fraction I factor out a z in the denominator...

I know cauchy reimann formulas, so I did it wrong, first off it doesn't satisfy c-r formulas, but for the imaginary part I should of taken the partial with respect to x as cauchy reimann implies, then that partial would be zero, then the derivative would be 6x^2 .

Homework Statement f(z)=2x^3+3iy^2 then it wants f '(x+ix^2) The Attempt at a Solution So I take the partial with respect to x and i get 6x^2 then partial with respect to y and I get 6iy, then I plug in x for the real part and x-squared for the imaginary part, then I get f '...

good point if we appraoch x from the left it goes to minus infinity , and if we approach from the right it is positive infinity, so the limit does not exist.

Homework Statement lim as z--> i , \frac{z^2-1}{z^2+1} The Attempt at a Solution [/B]When we plug in i we get -2/0, so we get division by 0, Does this mean the limit is infinity, I also tried approaching from z=x+i where x went to 0, you get the same answer, I also approached from...

I got it figured out, i just need to be more careful with the i and w . thanks for your help.

with the negative sign it goes around the circle in the opposite direction., but when I put the shift in the -2i translation, it makes the algebra more difficult, my teacher said using the modulus form of circle to make the algebra more clean and simple, but I can't quite get it to work...