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complex analysis



Need some help. Pretty sure I've done a) correctly, but I'm not sure how to begin with b). Any hints?
Reply 1
Avatar for davros
davros
16
Original post by Karaiyar


Need some help. Pretty sure I've done a) correctly, but I'm not sure how to begin with b). Any hints?


I'm quite rusty on this, but the first step would be to come up with an appropriate parameterisation for each segment of the path γ1\gamma_1 - can you do this?
Reply 2
Avatar for Hasufel
Hasufel
16
for the 1st one in b), you could try using the parametrization of the line segment from z0z_{0} to z1z_{1}

it being: z(t)=z0(1−t)+z1t,0≤t≤1z(t)= z_{0}(1-t)+z_{1}t, 0 \leq t \leq 1

so, putting z_0 = 2-2i, z_1=2+2i and simplifying you get:

f(z)=(2−2i)+4it,df=4idt f(z)=(2-2i)+4it, df = 4idt

the integral then becomes:

∫014i(f(z(t))−1−i)dt\displaystyle \int_{0}^{1} \frac{4i}{(f(z(t))-1-i)} dt

the f(z)-1-i simplifies it to:

∫014i(1−3i)+4it)dt\displaystyle \int_{0}^{1} \frac{4i}{(1-3i)+4it)} dt

and you can see this is of the form f ' (x) / f(x)

Amore efficient way is to say that it`s the line segment from y=-2 to y=2, and that you can call: z = 2+iy, -2<-y<=2 and you`ll be integrating wrt y

then, z' =idy,

f(z)= 1/((2+iy)-i-1)

so youre integral is:

∫−31i(1+(y−1)i)dy\displaystyle \int_{-3}^{1} \frac{i}{(1+(y-1)i)} dy

(someone please verify, i`ve done this in a rush, but i think it`s cool)
(edited 9 years ago)
Reply 3
Avatar for Karaiyar
Karaiyar
OP
Thank you!
Reply 4
Avatar for ztibor
ztibor
10
Original post by Hasufel
for the 1st one in b), you could try using the parametrization of the line segment from z0z_{0} to z1z_{1}

it being: z(t)=z0(1−t)+z1tz(t)= z_{0}(1-t)+z_{1}t

so, putting z_0 = 2-2i, z_1=2+2i and simplifying you get:

f(z)=(2−2i)+4it,df=4idt f(z)=(2-2i)+4it, df = 4idt

the integral then becomes:

∫014i(f(z(t))−1−i)dt\displaystyle \int_{0}^{1} \frac{4i}{(f(z(t))-1-i)} dt

the f(z)-1-i simplifies it to:

∫014i(1−3i)+4it)dt\displaystyle \int_{0}^{1} \frac{4i}{(1-3i)+4it)} dt

and you can see this is of the form f ' (x) / f(x)

Amore efficient way is to say that it`s the line segment from y=-2 to y=2, and that you can call: z = 2+iy, -2<-y<=2 and you`ll be integrating wrt y

then, z' =idy,

f(z)= 1/((2+iy)-i-1)

so youre integral is:

∫−31i(1+(y−1)i)dy\displaystyle \int_{-3}^{1} \frac{i}{(1+(y-1)i)} dy

(someone please verify, i`ve done this in a rush, but i think it`s cool)


for the 1st method

I would use the γ(t)\gamma (t) notation to the parametrized curve
and γ˙(t)\dot{\gamma} (t) for the derivative wrt t
So the f(z) should be the function and f(γ(t))f(\gamma (t)) the parametrized
form of that

∫014i(1−3i)+4it)dt=ln⁡1+i1−3i\displaystyle \int_{0}^{1} \frac{4i}{(1-3i)+4it)} dt=\ln \frac{1+i}{1-3i}

for the 2nd method

∫−31i(1+(y−1)i)dy=∫−31i(1−i)+iydy=ln⁡11−4i\displaystyle \int_{-3}^{1} \frac{i}{(1+(y-1)i)} dy=\int_{-3}^{1} \frac{i}{(1-i)+iy} dy=\ln \frac{1}{1-4i}

THey are not the same
THe lower and upper limit of integral should be -2 and 2 here as you wrote for y
(edited 9 years ago)
The line integrals over the straight paths are straightforward - just parametrise the integral over those line segments, which someone's already explained how to do.

For the circular arc, it might be a good idea to parametrise z=reiθz = re^{i \theta} - where you should determine rr from the information provided and set up appropriate limits for θ\theta.
(edited 9 years ago)
Reply 6
Avatar for Hasufel
Hasufel
16
Original post by ztibor
for the 1st method

I would use the γ(t)\gamma (t) notation to the parametrized curve
and γ˙(t)\dot{\gamma} (t) for the derivative wrt t
So the f(z) should be the function and f(γ(t))f(\gamma (t)) the parametrized
form of that

∫014i(1−3i)+4it)dt=ln⁡1+i1−3i\displaystyle \int_{0}^{1} \frac{4i}{(1-3i)+4it)} dt=\ln \frac{1+i}{1-3i}

for the 2nd method

∫−31i(1+(y−1)i)dy=∫−31i(1−i)+iydy=ln⁡11−4i\displaystyle \int_{-3}^{1} \frac{i}{(1+(y-1)i)} dy=\int_{-3}^{1} \frac{i}{(1-i)+iy} dy=\ln \frac{1}{1-4i}

THey are not the same
THe lower and upper limit of integral should be -2 and 2 here as you wrote for y


Thanks! - i suspected there may not be something quite right, as i`d hurried it.

I think for the first one, can`t it just be taken as cos(z) from the start point to the end point due to path independance since cos(z) is entire?

YUP (had a look) as i suapected, cos(z) is entire (analytic in the entire complex plane) and an integral from one point to another point is independant of the path chosen so the integral is just: ∫z1z2cos(z)dz \displaystyle \int_{z_1}^{z_{2}} cos(z) dz
(edited 9 years ago)

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