comp ana srj

8/10/2019 Comp Ana SRJ

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1. Dr. Prof. S.R. Jayaram

Department of Mathematics

SBM Jain College of Engineering,Jakkasandra 562112, Bangalore rural dt

(Please give your feed back to [email protected])

2.Complex analysis

Complex number

a +ib, i=-1

Exponential form

x+iy = r{cos-1

(y/x )+i sin-1

(y/x)}, r = (x2

+y2

)

Eulers formula

eix = cosx + isinx

Representation of complex number

-Argonds plane, Complex plane


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5.Complex numbers

Adding (a+ib) (c+id) = (a+c) i(c+d)

Multiplication

(a+ ib) x (c + id) = (ac - bd) i(bc+ad)

Division

(a + ib)/(c + id)

= [(a + ib)/(c + id)]x (c-id/c-id)

= [ac+bd) +i(bc-ad)]/[c2

+ d2

]

6. Complex numbers are not ordered

Equality (a+ib) = (c+id) iff a =c and b =d

i= -1, i2 = -1, i3 = -i, i4=1

1/i =-i, 1/i2 = -1,1/ i3 =-i, 1/i4=1

In exponential form only principal value or periodic function is taken

7 Functions- complex

Single valued

f(z) = f(x + iy), i = -1

Differentiability

CR equations ux

=vy

, uy

= -vx

f(z) = ux + ivx or -iuy - vx

f(z) is analytic

f(z) = Logz, ez, z+1/z, z

2,

1/2log(x2+y

2) +itan

-1y/x


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11. Complex analysis

If u(x,y) = 2xy + xy2 -2y3, is u harmonic?

Ux =2y + y2

, Uxx

= 0

Uy =2x + 2xy -6y2

Uyy

= 2x12y

Uxx+ Uyy 0, so U is not harmonic

12.Complex Analysis

Show V = -sinx sinhy is harmonic. Find its conjugate U

Answer: Vx = -cosx sinhy, Vxx = sinx sinhy, Vy = -sinx coshy, Vyy = -sinx sinhy

So Vxx + Vyy = 0

From CR equations, Ux = Vy and Uy = -Vx

So u(x,y) = cosx coshy + constant = -Vx

So u(x,y) = cosx coshy

f(z) = cosx coshy +c +i(-sinx sinhy)

13. Complex Analysis

Find analytic function, given

u = ex

(xcosy-ysiny) + 2sinxsinhy+ x3

- 3xy2

+ y

Ux = ex(xcosy-ysiny) + excosy +

2cosx sinhy + 3x23y2

Uy =ex

(-xsiny-sinyy cosy) + 2sinx coshy

-6xy +1

f(z) = Ux + i Vx = Ux - i Uy = zez+e

z+3z

2-2isinz-i replacing x by z and y by zero)

Integrating, f(z)= zez+ez+z3/3 -2icosz-iz +c


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If V = (x2y2 )+ x/(x2 + y2 ), find u

V is harmonic, Vxx

+ Vyy

= 0

Vx = -2y2xy/(x2

+ y2

)2,

= Uy

Integrating U = -2xy + y/ (x2 + y2 ) +c(y), where c(y) is a function of y

Now u = -2xy + y/ (x2 + y2 ) +c

15 Complex analysis

Given u = sin2x/(cosh2ycos2x) find v

Let f(z) = u + iv

f(z) = ux

+ i vx

= vx

- i uy

= (2 cos 2x cosh 2y2 )/ (cosh2ycos2x)2

- i (2 sin 2x sinh 2y )/ (cosh2ycos2x)2

=By Milne Thompson method,

f(z)= (2cos2z-2)/ (1-cos2z)2 + i(0)

= -cosec2z. So f(z) = cot z + c


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16.Complex analysis

Find the analytic function given

v = r2

cos 2 r cos + 2

r ur = v = -2 r 2 sin 2 + r sin

-1/r u = vr = 2r cos - cos , so

ur

= -2r sin 2 + sin , Integrating w.r.t r

u = -r2

sin 2 + rsin + c

So f(z) = i(r2

e2ir ei) + c + 2i

17 Complex analysis

If f(z) is regular, prove

(uxx + uyy)|f(z)|2 = 4 |f(z)|2

f(z) = u + iv, so that |f(z)|2 = u2 + v2 = (x,y)

x = 2uux + 2vvx and

xx = 2[ uuxx + (ux)2

+ vvxx +(vx)2

], similarly

yy = 2[ uuyy + (uy)2

+ vvyy +(vy)2

],

So xx+yy= 2[u(uxx+uyy ) + v(vxx + vyy)]+ 2[(ux )2

+(u y)2]+ (vx )

2+(vy)

2]

By CR equations, (ux )

2

= (vy )

2

, (u y)

2

= (-vx )

2

And uxx+uyy = 0, vxx+vyy = 0,

so xx+yy= 4[(ux )2

+(u y)2 ] = 4|f(z)|2


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Find if uv = (cosx + sinxe-y)/2(cosxcoshy), f(/2)= 0

Ux- Vx= [(sinx-cosx)coshy + 1e-y sinx]/

2[cosxcoshy)2 (1)

Uy- Vy= [(cosx - coshy)e-y

+

(cosx + sinxe-y )sinhy]/2[cosxcoshy)2

-Vx- Ux= [(sinx + cosx)sinhy + e

-y(cosxcoshy -sinhy]/2[cosxcoshy)2

(2)

Subtracting (2) from (1) and adding (2) and (1)

And putting x = z, y = 0, we get

19. Complex Analysis

f(z) = Ux

+ iVx

= (1- cos z) /2(1- cos z)2

Integrating, f(z) = -1/2 cot (z/2) + c Using initial condition that f(/2) = 0

We get c =

So f(z) = ( 1cot z/2)


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20 Complex Analysis

Given v = (r -1/r)sin, find u

CR equations are u

r

= 1/r v

and u

= -r v

r u = -r vr = -(r + 1/r) sin , Integrating

u = (r + 1/r) cos + c(r), Differentiating

Using CR c(r) = 0, so u = (r + 1/r) cos

f(z) = (r + 1/r) cos + i(r - 1/r) sin + c

21 Transformations

1. Translation w = z + c

Taking z = (x + iy) and c = (a + ib)

w = c + z = (x + a) + i( y + b), so a point

(x, y) is mapped onto (x + a) + i( y + b).

The object is moved through a distance c.

Image retains same shape and size


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24 CA-Inversion and reflection

W = 1/z

Let z = re

iand w = Re

i

Then R = 1/r and = -

25 . CA-Bilinear transformation

W = (az + b)/(cz + d), a, b, c, d are complex constants and ad - bc 0. The

condition ad - bc 0 implies dw/dz 0.

The inverse transformation is given by

z = (-dw + b) / (cw - a)Fixed points of transformation is the solution

of cz2

+ (d-a) zb 0

26. CA-bilinear transformation

By actual division w = (az + b)/( cz + d) is

w = a/c + (bcad)/c2

. 1/(z+d/c)So consider w1 = z+d/c, w2 _= 1/w1

And w3

= (bc - ad)/c2

w2

w = a/c + w3

Translation, rotation, magnification


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27 CA- Bilinear transformation

The cross ratio of four points z1 , z2 , z3 , z4

is denoted by (z1 , z2 , z3 , z4 ) and is defined by (z1 - z2 )(z3 - z4)/ (z1 - z4 )(z3 - z2)

Bilinear transformation preserves cross-ratio of any four points

w1 = (az1 + b)/(cz1 +d), w2 = (az2 + b)/(cz2 d),

w3

= (az3

+ b)/(cz3

+d), w4

= (az4

+ b)/(cz4

+d),

28 CA-BLT

(z1

- z2

)(z3

- z4)/ (z

1- z

4)(z

3- z

2) =

(w1

- w2

)(w3

- w4

)/ (w1

-w4

)(w3

- w2)

So the bilinear transformation preserves cross ratios of any four points

29 Complex Analysis

Let us denote four complex constants a, b, c, d and define the function

f(z) = w = (az + b)/(cz+d) is called Mobius/bilinear transformation

There exists a unique transformation which maps three points z1,

z2,

z3 to the

points

w1

,w

2,w

3given by the formula

(z- z1)(z2-z3)/(z-z3)(z2-z1) =

( w - w1)(w2 - w3)/(w - w3)(w2 - w1)


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30 CA-Bilinear transformation

(-1, 1, i) to (-1, 0, 1)

W = i(1-z)/(1+z)

(-2, -1-i, 0 ) to (-1, 0, 1)

W =[(1- i)z+2]/[(1+i)z+2]

(1, i, -1) to (i, 0, -i)

W = (1+iz)/(1-iz)

(1, i, -1) to (i, 0, -i)

W = z(6+3i)+(8+4i)/ z(7+6i)+(6-2i)

31 CA-Bilinear transformation

Find the invariant points of the BLT

w = (z-1)/(z+1), z =i

W = (6z-9)/z, z = 3 Find BLT which maps (-1, 0,1) to (0, i, 3i)

w = -3i(z+1)/(z-3),

Find BLT which maps (1, 0,-1) to (i, 1, )

w = (-1+2i)z+1)/(z+1)

Find BLT which maps (0, 1, ) to (-1,- i, 1)

w = (z-i)/(z+i)


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32 Complex Analysis-Integration

Line integrals

c (u+iv)(dx+idy) =

c (udx-vdy) + ic (vdx+udy)

Properties of line integrals

Linearity

ac f(z)dz + bc g(z)dz =c (af(z)+ bg(z))dz

Sign reversalc f(z)dz = -c f(z)dz, where the c traversed in opposite direction

Partitioning of path c f(z)dz =c1 f(z)dz +c2 f(z)dz

33 Line integrals

Evaluatec |z|2dz around the square with vertices (0,0), (1,0), (1,1), (0,1)

The square is enclosed by four lines c1, c2, c3, c4


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34 Line integrals

cf(z)dz=c1f(z)dz +c2f(z)dz +cf(z)dz +c f(z)dz

Along c1, y=0,c1 |z|2dz =c1 (x2+y2)(dx+idy) =01 x2 dx = x3/3 = 1/3

Along c2, x = 1, y varies from 0 to 1

01

(1+y2

)idy = i4/3

Along c3, y =1, and x varies from 1 to 0

10

(x2

+1)dx = -4/3

Along c4, x=0, y varies from 1 to 0

10 y2 idy = iy3/3 = -i/3

cf(z)dz = -1+i

35 Line integrals

c dz/(z-a) = 2i where c is the circle |z-a| = r

The parametric equation for C is z-a= rei

where varies from 0 to 2

dz = irei d

c

dz/(z-a) =c

(1/ rei

)irei

d = 2i


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36 Line integrals

Evaluate (x2

+iy)dz, along the paths y =x and y =x

2

for z varying from 0 to 1+ i

dz = dx + idy, so (x2 +iy)(dx + idy) =

Along y = x, (y2 +iy)(dy +idy) = (1+i)(y3/3+iy2/2)

For y varying from 0 to 1

= (1+i) (1/3 + i/2)

= (1/3 -1/2) + i(1/3 + ) = -1/6+i/6

37 Line integrals

If f(z) is analytic function and f(z) is continuous at each point within and on a

closed curve C then

c f(z)dz = 0

Writingc f(z)dz =c (udxvdy) + ic (vdx + udy)

Since f(z) is continuous ux

uy

vx ,

vy

are continuous in the region D enclosed by

C, Using Greens theorem,

c f(z)dz = -D [vx + uy] dxdy + i-D[ux -vy]dxdy. Since f(z) is analytic,

vx = - uy ,ux = vy and the two double integrals vanish identically.


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38 Line integrals

If f(z) is analytic function and if a is any point within C,

then f(a) = 1/2ic f(z)/(z-a)dz

Proof: Function f(z)/(z-a) is analytic, at all points of C except at z = a, We draw a small

circle C1 of radius r lying entirely within C. Now f(z)/(z-a) is analytic in the region

enclosed by C and C1 , by Cauchys theoremc f(z)/(z-a)dz =c1f(z)/(z-a)dz

For any point on C1 , z-a = rei

and dz = irei

d

c1f(z)/(z-a)dz =c1f(a+ rei )/ rei .i rei d

= ic1f(a+ rei )d . In the limiting case as r0,

ic1f(a)d= 2i, thus f(a) = 1/2ic f(z)/(z-a)dz

39 Similarly f(a) = 1/2ic f(z)/(z-a)2dz ,

f(a) = 2!/2ic f(z)/(z-a)3dz ,

f(a) = 3!/2ic f(z)/(z-a)4dz

.

fn(a) = n!/2ic f(z)/(z-a)n+1dz


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40 Cauchys integral formula

Evaluatec(z2

z+1)/(z-1) dz where C is the circle1. |z| = 1.

Answer: Given f(z) = (z2z+1) and a =1.

Since f(z) is analytic within and on |z| = 1and

a = 1 lies on C. By Cauchys integral formula

1/2ic f(z)/(z-1)dz = f(a) =1

2. |z| = . Lies inside the circle |z| = 1. Hence by

Cauchys theoremc(z2z+1)/(z-1) dz =0.


c(e2z

/(z-1)(z-2) dz, where C is the circle |z| = 3

Answer: f(z) = e2z

is analytic within the circle |z|=3,

a = 1 and a = 2 lie inside |z| = 3. Hence

c(e

2z/(z-1)(z-2) dz =

c(e

2z[1/(z-2) -1/ (z-1)] dz

=c(e

2z .1/(z-2)dz -

c(e

2z1/(z-1)dz

= 2ie4 - 2ie2 = 2ie2 (e2 -1)


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c(cosz/(z2

-1)dz, around a rectangle 2 i, -2 i

Answer: f(z) = cos z is analytic in the rectangle and the points z = 1, and z = -1, lieinside the rectangle.

c(cosz/(z2 -1)dz =

c[(cosz/(z -1)dz -(cosz/(z+1)dz ]

= [ 2icos(1) - 2icos(-1)]

= 0.


Taylors series: If f(z) is analytic in a circle C with center a, then for z inside C

f(z) = f(a) + f(a) (z-a)+ f(a) (z-a)2 /2! +..

+ +fn (a) (z-a)n /n! +.

Laurents Series: If f(z) is analytic in a ring

shaped region R bound by two concentric circles

C1 and C2 of radii r and r1 with r > r1 center a,

then for all z in R

f(z) = a0 + a1(z-a) + a2(z-a)2 +..+ a-1(z-a)

-1+

+ a-2(z-a)-2 +..+ where an = 1/2if(t)/(t-a)n+1 dt


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44 Taylors, Laurent Series

1. Find Taylors series expansion of 1/(z+1)2 around z = -i

Answer: f(z) = 1/(ti+1)2 = (1i)2 [ 1 + t/(1-i)]-2

= i/2[ 1- 2t/(1-i) + 3t2/(1-i)24t3/(1-i)3 + ..]

2. Expand f(z) = (2z3

+1)/z2+z) about z = i

Answer: f(z) = (2z3

+1)/z2+z) = (2z-2) + (2z+1)/z(z+1)

(2i-2) +2(z-i) + 1/z + 1/z+1), Put zi = t

1/z = 1/(t+i) = 1/i(1+t/i)-1 = 1/i[1- t/I + t2/i3 - t3/i4 +..]

= -i + ( zi) +2 (-1)n(z - i)

n/t

n+1)

45 T, L series

1/(z+1) = 1/(t+i+1)

= 1/(i+1)[1+ t/(1+i)]-1

= 1/(1+i) [ 1- t/(1+i) + t2/(1+i)

2t3/(1+i)3 + -..

= (1-i)/2t/2i + t2/(1+i)3t3/(1+i)4+- ..

= - i/2(z - i)/2i +2 (-1)n(z-i)

n/(1+it)

n+1

Substituting, f(z) = (i/3 - 3/2) + (3 + i/2)(z-i) +

2 (-1)n [1/in+1 +1/(1+i)n+1 (z-i)n


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46 TL Series

2. Expand f(z) = 1/(z-1)(z-2) in the region

a. |z| < 1

Answer: Resolving into partial fractionsf(z) = 1/(z-1)(z-2) = 1/(z-2)1/(z-1)

= -1/2 (1- z/2)-1

+(1 - z)-1

For |z| < 1 both |z/2| and |z| are < 1

So -1/2 (1- z/2)-1

+(1 - z)-1

=

= -(1+z/2+z2/4 + z

3/8+.+(1+z+z2+z3+.)

= + z+ 7/8 z2

+ 15/16 z3 + .

47 TL Series

Expand f(z) = 1/(z-1)(z-2) in the region

for 1 < |z| < 2

Answer: Writing f(z) as - .1/(1-z/2)1/z(1-z-1)-1

Both |z/2| and |z-1

| are less than 1. hence

= - (1+z/2 +z2/4 + z

3/8 + .) 1/z(1+ z-1 + z-2+ z-3 + z-4 + ..)

= - z-4 - z-3 - z-2 - z-1 -1/4 z -1/8 z2- .

Which is a Laurent series


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48 TL Series


|z| > 2

Answer: For |z| > 2,

f(z) = 1/z(1 - 2z-1

)1/z(1 - z-1)-1

= z-1

( 1 + 2z-1

+ 4z-2

+ 8z-3 + ..) -z-1(1 + z-1 + Z-2 + Z-3+..)

= .+ 8Z-4

+ 4Z-3

+ 2Z-2

- 1 - Z - Z2

49 TLSeries


0< |z1 | < 1.

Answer: Writing f(z) = 1/(z-1)(z-2) = 1/[(z-1)-1]1/(z-1)

= -(z-1)-1[1-(z-1)]-1

= -(z-1)-1

- [ 1+ (z-1) +(z-1)2

+ (z-1)3 .. ]


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50 TL Series

5. Expand f(z) = z/(z-1)(z-3) in power series in

a. 1< |z | < 3.

Answer: Resolving into partial fractions

f(z) = z/(z-1)(z-3) = -1/2/(z-1)+ 3/2/(z-3)

= - 1/z(1-1/z) + 3/2 1/-3(1-z/3)

= -1/2z(1-1/z)-1 (1 - z/3)-1

= -1/2z( 1+1/z+ 1/z2 + +..) ( 1+ z/3 + z2/9 + z3/27+..)

51 CA

6. Expand f(z) = z/(z-1)(z-3) in power series in

|z1 | < 2.

Answer: f(z) = -1/2/(z-1)+ 3/2/(z-3),

Put z = u +1, = - . 1/u + 3/2 . 1/(u+1-3) =

= - 1/(z-1) + 3/2u[ 1 - 2/u]

-1

=

= - 1/(z-1) + 3/2(z-1) +3/(z-1)2

+ 6/(z-1)3 +

52 Singularities

Zero: The value of z for which f(z) = 0, is called zero of f(z)

A point is a singular point if it ceases to be

Analytic at that point. If f(z) is analytic everywhere except for the point z = a, then z = a is

Called isolated singularity.

In f(z)=1/(z-a)n

, z = a is called a pole of order n.


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53 CA

1. In (zsinz)/z2

, z = 0, is a removableSingularity, since

(zsinz)/z2 = 1/z2 [ z - {zz3/3! +z5/5! - + }]

i.e. no negative powers of z are in the expansion.

2. Poles of 1/(1 - ez) are the solutions of (1- e

z) = 0 or

z = 2ni, n = 0,1,2,3, ..

Residue: The coefficient of (z-a)-1

, in the expansion of

f(z) around an isolated singularity is the residue of f(z) at

that point.

54 Residue theorem

Residue f(a) = 1/2icf(z)dz

that iscf(z)dz = 2i f(a)

1. If f(z) has a simple pole at z = a,

then Resf(a) = Lt [ (z-a) f(z)

za2. If f(z) has a pole of order n at z = a, then

Res f(a) = 1/(n-1)! [ dn-1

/dzn-1

(z-a)n

f(z)] at z = a


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55 CA

Determine poles of f(z) = z

2

/(z-1)

2

(z+2) and residue at each pole, hence evaluatecf(z)dz

Where c is the circle |z| = 5/2

Answer: Limit as z -2, of (z+2)f(z) = 4/9

Simple pole at z = -2 and residue = 4/9

Limit z1, of (z-1)2 f(z), -1 is a pole of order 2,

Res f(1) = d/dz(z-1)2f(z) at z = 1, = 5/9

cf(z)dz = 2i(sum of the residues) = 2i(4/9+5/9)

56 CA-Residues

Evaluatec(z-3)/(z2 +2z +5)dz, where c is the circle |z|=1

Answer: Solving z2 +2z +5 = 0,

z = -1+2i, -1-2i. Both the

Poles of f(z) lie outside the circle |z| = 1,

hence (z-3)/(z2

+2z +5) is analytic in |z| = 1

and so by Cauchys theorem

c(z-3)/(z2 +2z +5)dz = 0.


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57 CA-Residues-Integrals

Evaluatec(z-3)/(z2

+2z +5)dz, where c is the circle

|z+1-i|=2.

Answer: the poles of (z-3)/(z2 +2z +5) are -1+2i

and -1-2i. The pole -1+2i lies inside the circle

|z + 1 - i| = 2. So f(z) is analytic except at this point

c(z-3)/(z2 +2z +5)dz = Resf(-1+2i).

So limit z -1+2i of [z-(-1+2i)f(z)] = +i

Hence by Residue theoremc(z-3)/(z2

+2z +5)dz

= 2i(i + ) =(i - 2)


Evaluatec(z-3)/(z2

+2z +5)dz, where c is the circle

|z + 1 + i| = 2.

Answer: the poles of (z-3)/(z2 +2z +5) are -1+2i

and -1-2i. The pole -1- 2i lies inside the circle

|z + 1 + i| = 2. So f(z) is analytic except at this point

c(z-3)/(z2

+ 2z +5)dz = Resf(-1 - 2i).

So limit z -1 - 2i of [z-(-1 + 2i)f(z)] = -i

Hence by Residue theoremc(z-3)/(z2 +2z +5)dz

= 2i(-i + ) =(i + 2)


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Evaluatec (tanz) dz, where c is the circle |z | = 2.

Answer: The poles of tanz = sinz/cosz are at

z = (2n+1)/2, n = 0,1,2,3. Of these only z = -/2,/2, lie inside the circle |z| = 2. So f(z) is analytic except

at this point, soc(tanz)dz = 2i[Resf(/2) + Resf(-/2)]

Res f(/2) = So limit z /2 sinz/-sinz = -1

Res f(-/2) = So limit z -/2 sinz/-sinz = -1

Hence by Residue theoremc(tanz)dz = 2i(-1 - 1) = -4i


Evaluatec (sinz2

+ cosz2)/(z-1)2 (z -2) dz, where

c is the circle |z| = 3.

Answer: The pole z = 1 is of second order and z = 1 is a

simple pole. So f(z) is analytic except at these points.

c

(sinz2 + cosz2)/(z-1)2 (z -2) dz = Sum of Residues at

1 and 2. Residue at z = 1,

= d/dz [(sinz2 + cosz2)/(z -2)]= 2 + 1

Residue at z = 2, Lt z 2, (sinz2 + cosz2)/(z-1)2 = 1

c (sinz2

+ cosz2)/(z-1)2 (z -2) dz = 2i(2 + 1 + 1) = 4( +1)i

comp ana srj

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