The value of the integral - ccos2- z2z-1z-3dz Where C is a closed curve given by -z-1 is

∫ _ccos2π z(2z 1)(z 3)dz=∫_Cf(z) dz Where (C):|Z|=1 Only one pole z=1/2 lies inside the circle nbsp;C Res (f(z) at z=1/4) =lim_z=→12{ ( z 22 )f(z) } =lim_z=→1 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Other

Engineering Mathematics

Residue Theorem

The value of ...

Question

The value of the integral $\int_{c} \frac{c o s 2 π z}{(2 z - 1) (z - 3)} d z$ (Where $C$ is a closed curve given by $| z | = 1$ is

i

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{π i}{5}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{2 π i}{5}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

π i

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $\frac{2 π i}{5}$
$\int_{c} \frac{c o s 2 π z}{(2 z - 1) (z - 3)} d z = \int_{C} f (z) d z$
Where $(C) : | Z | = 1$
Only one pole $z = 1 / 2$ lies inside the circle $C$
Res ( $f (z)$ at $z = 1 / 4$ )
$= {lim}_{z =\to \frac{1}{2}} {(z - \frac{2}{2}) f (z)}$
$= {lim}_{z =\to \frac{1}{2}} {\frac{c o s (2 π z)}{2 (z - 3)}}$
$= \frac{c o s (2 π \frac{1}{2})}{2 (\frac{1}{2} - 3)} = \frac{1}{5}$
By Cauchy's residue's theorem,
$I = (2 π i) (\frac{1}{5}) = \frac{2 π i}{5}$

Suggest Corrections

Similar questions

Q. The value of

\oint_{C} \frac{s i n z}{z} d z,

where the contour of the integration is a simple closed curve around the origin is

C

is a closed path in the

z -

plane given by

| z | = 3

. The value of the integral

\oint_{C} (\frac{z^{2} - z + 4 j}{z + 2 j}) d z

Q. The value of the integral is

(\frac{2 z + 1}{z^{2} + 1}) d z

where

c

| z | = \frac{1}{2}

Q. The value of

\oint_{r} \frac{3 z - 5}{(z - 1) (z - 2)} d z

along a closed path

Γ

is equal ot

4 π i

, where

z = x + i y

and

i = \sqrt{- 1}

.
The correct

Γ

Q. Many ROCs of are possible, ROC for which the given sequence is a casual sequence is

X (z) = \frac{z (z - 4)}{(z - 1) (z - 2) (z - 3)}

| z | > a

where

a

Join BYJU'S Learning Program

The value of the integral ∫ccos2πz(2z−1)(z−3)dz (Where C is a closed curve given by |z|=1 is

The correct option is C 2πi5∫ccos2πz(2z−1)(z−3)dz=∫Cf(z)dz Where (C):|Z|=1 Only one pole z=1/2 lies inside the circle C Res (f(z) at z=1/4) =limz=→12{(z−22)f(z)} =limz=→12{cos(2πz)2(z−3)} =cos(2π12)2(12−3)=15 By Cauchy's residue's theorem, I=(2πi)(15)=2πi5

The value of the integral $\int_{c} \frac{c o s 2 π z}{(2 z - 1) (z - 3)} d z$ (Where $C$ is a closed curve given by $| z | = 1$ is