Question

Find the integral of (sin(pi*z^2)+cos(pi*z^2))/(z-1)(z-2) with respect to absolute z=3.

and Find the integral of (e^2z)/(z+1)^4 with respect to absolute z=3.

Answer 1

you mean radius 3 ??

Answer 2

|z| = 3?

Answer 3

Yes.

Answer 4

make use of residue theorem ... this case both z=1,2 lies both inside the radii |z| = 3

Answer 5

the first one is easy .. .i guess.

Answer 6

for second one ...
\[ \lim_{z \rightarrow -1} {1 \over 4!} {d\over dz}(z+1)^4{e^{2z}\over (z+1)^4}\]

Answer 7

Oops!!
\[ \lim_{z \rightarrow -1} {1 \over 4!} {d^3\over dz^3}(z+1)^4{e^{2z}\over (z+1)^4}\]

Answer 8

that would be the residue for second. The itegral would be 2 \pi i Residue( f(z) )
which you can calculate as above.

Answer 9

WIth the problem of sin(z) =2, the steps I followed was to express (sin (x)= (e^z-e^(-1))/2i = 2 and (e^z-e^(-1) = 4i ending up with e^x * sin y =2. Is that the correct way to go or is there an alternative.

Answer 10

(e^z-e^(-z) = 4i
(e^z)^2 - 4i e^z- 1 = 0
x^2 - 4i x - 1 = 0
x = 4i +- sqrt((4i)^2 + 4))/2
e^z = ....
z = ln(....)

Answer 11

Many thanks. In the proof that all values of (1-i)^(isqrt{2}) lie on a straight line can you assist with the steps to follow?

Answer 12

follow this
http://math.stackexchange.com/questions/189703/does-ii-and-i1-over-e-have-more-than-one-root-in-0-2-pi

Answer 13

if it has more than one root in [0, 2pi] then it should be false. If it has only one root in .. then [0, 2pi], then it would be branches of it ... which would mean that they all lie in straight line.

Answer 14

I am not so sure .... this is just mu hunch ... you better try asking people on http://math.stackexchange.com