Gaziantep Univ., Dept. Electrical & Electronics Eng., Complex Variables Midterm-II Exam 22 May 2013 Duration: 100 Min.

2C 1C

a

b

g

h

d k

l

0z

e

Number: Name & Surname: Sign: PLEASE USE THE BACK FACE OF THIS PAPER AND ANY ADDITIONAL PAPER(S) FOR YOUR ANSWERS

Q1 (20 points) For the region defined by two contours ( 1C and 2C ) below, determine ( )( )0

nc

f zdz

z z−∫ where C is

any contour containing the point 0z [Please utilize the derivation given in the lecture]. A1) From the configuration of the problem, we obtain

( )( )

( )( )

( )( )

( )( )

( ) ( k)

0 0 0 00

e via g b viad a

n n n na e d b

f z f z f z f zdz dz dz dz

z z z z z z z z+ + + =

− − − −∫ ∫ ∫ ∫ (4P) (1)

( )( )

( )( )

( )( )

( )( )

( )( )

( ) ( )

0 0 0 0( 1)

02 .1 !

a via h d via lb e

n n n ne a b d

n

f z f z f z f zdz dz dz dz

z z z z z z z z

f zi

nπ

−

+ + +− − − −

= −−

∫ ∫ ∫ ∫ (4P) (2)

Summation of Eqs. (1) and (2) yields, after some manipulations,

( )( )

( )( )

( )( )

1 2

( 1)0

0 02 .

1 !

n

n nC C

f zf z f zdz dz i

nz z z zπ

−+ =

−− −∫ ∫ (2P) (3)

Q2) (30 points) Evaluate the integral ( )( )2cot

2C

zdz

z i

π

−∫ over the contour inside the region bounded by 1 :C 0.5z =

(ccw) and the contour over the square with center 0.2cz = − and side length of 0.9 (cw). A2) Let us firstly analyze the analyticity of the integrand. At 2z i= and 0, 1, 2,...z = ∓ ∓ , the integrand is not analytic. For the region bounded by the contours, none of these points is inside the region. Therefore, the integration is zero according to the Cauchy integral theorem.

UGaziantep Univ., Dept. Electrical & Electronics Eng., CComplex Variables Midterm-II Exam 22 May 2013 Duration: 100 Min.

Q3) (a) Find the Laurent series of the function ( ) ( ) ( )( ) ( )2

3 2 6 17

4 6 9 12

i z if z

z i z i

+ + −=

− + − − for 0 0z = and 3 4z< < (30

points), and (b) Find its Taylor series if any (explain your answer). (a) The function ( )f z can be expressed by

( ) ( ) ( )( ) ( )

( )( ) ( )

( ) ( ) ( )1 22 23 2 6 17 2 2 4 3 3 9 2 3 .

3 4 34 6 9 12 4 6 9 12

i z i zi i i z i if z f z f zz i z iz i z i z i z i

+ + − − + + −= = = + = +

− − +− + − − − + − − (8P) (4)

Let us analyze ( )1f z first. The point 3i is the decision-making point. For the region 3 4z< < , we have

( ) ( ) ( )1 11 0 0

2 2 2 1 2 2 3 ,3 131

n n n

n n

i i i i if z q i zz i z q z ziz

z

∞ ∞−

= == = = = =

− −⎛ ⎞⎛ ⎞− ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

∑ ∑ (8P) (5)

where

1 13 , 1, 3.iq q zz

= < > (6)

For the function ( )2f z and the given region 3 4z< < , we obtain

( )( ) ( )

( ) ( ) ( ) ( )2 22 0 0

2 2 1 2 2 ,4 3 1 4 3 4 3 4 3

4 3 14 3

nn

n n

i i i i zf z qi q i i izi

i

∞ ∞

= =

⎛ ⎞= = − = − = − ⎜ ⎟+ − + + +⎛ ⎞ ⎝ ⎠− + −⎜ ⎟⎜ ⎟+⎝ ⎠

∑ ∑ (8P) (7)

where

( )2 2, 1, 5.4 3

zq q zi

= < <+

(8)

Therefore the solution will be the summation of ( )1f z and ( )2f z in Eqs. (5)-(8) (2P). (b) Since Eq. (5) contains a term of z ( 0 0z = ) with a negative power of -1, the solution has no Taylor series expansion (4P). Q4) Find (a) the center of the series (5P) and (b) the constant u value (15P) if the radius of convergence of the

series ( )( )( ) ( )5

0

22 3

1

nn

nn

u iz i

i

∞

+=

+− +

+∑ is equal to 2.

(a) The center of the series from the format ( )0nz z− is 0 3 2z i= − + (5P).

(b) The radius of convergence of the series is

( )( )( )

( )( )

( )( )( )

( )( )

6

151

2 1 1 12,

2 221

n nn

nnn n nn

u i i i iaR Lim Lim Lim

a u i u iu ii

+

++→∞ →∞ →∞+

+ + + += = = = =

+ +++ (10P) (9)

1 2, 2 1, 3 .uu i u i i+ = = − = − (5P) (10)