d)if g(z)=z^2*e^z and y(t) is the path in part c, find an upper bound for
|∫g(z)dz|
y(t)=i+3e^2it [0,pi/4]
i'm trying to figure out if i have done this right, Usually in other examples the
3e^2it is on its own, that is centred at zero, and i can use the fact that the magnitude will simply be the radius( three). With this example can i do that, or do i have to seperate the imagineray and real parts?
doing it like this seems quite involved:Anyway am i "overcomplicating it as i usually do! or is this the right approach.For the magnitude of the path i get:If anyone has done this type of problem i could use some help.
[(3cos2t)^2+(1+3sin2t)^2]^1/2
=[9(cos2t)^2+1+6sin2t+9(sin2t)^2]^1/2
=[10+6sin2t]^1/2
or should i use:
|z-i|=3
L=R*pi/4=3pi/4
then g(z)=z^2*e^z
|∫g(z)dz|=|z^2*e^z|<=|z^2||e^x|=
(then using the fact that the maximun of cos and sine is one.)
=|18(cos2t)^-9+sin4t||e^3cos2t|
<=|10||e^3|
therefore:
|∫g(z)dz|<=(e^3)30pi/4 =e^3(15pi)/2