MATH 656 S11 Midterm Solutions

8/2/2019 MATH 656 S11 Midterm Solutions

1/6

x

ln(2+31/2

2 2

ln(2+31/2)

y

/2

Math 656 March 10, 2011

Midterm Examination Solutions

1) (14pts) Derive the expression for sinh1 z (arcsinhz) using the definition of sinh w in terms ofexponentials, and use it to find all values of sinh

1(2i). Plot these values as points in the complex

plane. Make sure your points agree with the period of the hyperbolic sine function.

{ }

{ } { }

2

2

2 2 1/ 2 1 2 1/ 2

1 1/ 2

sinh 2 0 2 1 02

2 1 0 ( 1) sinh log ( 1)

sinh (2 ) log 2 ( 4 1) log 2 3 ln(2 3) 2 ln(2 3) 22 2

w ww w z w w

s s

e ez w e e z e e w e

s zs s z z w z z z

i i i i i i k i i

= = = =

= = + + = = + +

= + + = = + + = + + +

This agrees with the period of hyperbolic sine function, which is 2i


2/6

u

vw = cosz

y

x

/2/2

z

2) (15pts) Show that images of vertical lines under transformation w = cosz are hyperbolic2 2

2 21

u v

a b

=

. What shape are images of horizontal lines? What is the angle between these two sets

of curves?

2 22 2

2 2

cos cos( ) cos( ) cos( ) sin( )sin( ) cos cosh sin sinh

, ,

cos cosh sin sinh

cosh sinh 1: 1cos sin

1. Image of vertical lines: is fixed

Use identity

u v

w z x iy x iy x iy x y i x y

z c iy y c

w c y i c y

u vy y

c c

= = + = =

= + < <

=

= =

cos 0 sin 0)

cos 0, sin 0, ( , 1) (1,

A hyperbola (unless or

If the image is the v-axis (see figure). When the images are real: or

c c

c c

= =

= =

2 22 2

2 2

, ,

cos cosh sin sinh

cos sin 1: 1cosh sinh

2. Image of horizontal lines: is fixed

Use identity An ellipse (or a line [-1,1] if sinh c = 0)

u v

z x ic x c

w x c i x c

u vx x

c c

= + < <

=

+ = + =

This is a conformal mapping, which preserves angles, so images of horizontal and verticallines are orthogonal wherever (cos ) ' 0z , since the horizontal and vertical lines are

orthogonal.


3/6

3. (16pts) Is the functionf(z) = /z z continuous for allz? Is it differentiable anywhere? Is it analytic

anywhere? Prove your answers directly (using limits), and verify your answer about analyticity using

Cauchy-Riemann equations.

2 2 2

2 2 2 2 2

2( ) /

( )

2( )

| |

R

cos(2 ) sin(2 )

Im

)

e

is continuous wherever its real and imaginary parts are c

(or, in polar form

ontinuous

and a

i

v

i

u

i

u v

f z

z z x y xyf z i

z z x y x y

f

f z re r i

f

e e

= = = +

+ +

= = = +

0, ( )

0

re continuous apart from point where is not even defined,

so the only discontinuity is

x y f z

z

= =

=

Now lets examine differentiability:

The simplest way to show non-analyticity in this case is to differentiate in polar coordinates(see homework #4), but we can also use the usual Cartesian representation:

0 0 0 0

0

( ) ( ) 1 ( ) ( )lim lim lim lim

( ) ( )

0

lim(

Let's take the derivative along the horizontal directions:

h h h h

x

z h z hdf f z h f z z h z z z h z z h

dz h h zz h h z z h h z z h

h h x

df x z z

dz x z z

+ + + + = = = =

+ + + = =

=

2 2

2 20

Im2

) ( ) ( )

, , 0

Relim 2

( ) ( ) ( )

Now let's differentiate along the vertical direction:

These two expressions are never equal (note that

y

z z zi

x z z

h i y h i y y

i y z zdf z z z

dz i y z z i y z z

= =

+

= =

+ += = =

)z=0 is outside of domain of definition

Also, lets check Cauchy-Riemann equations:

( ) ( )

( ) ( )

2 2 2 2 2 2 2

2 2 2 22 2 2 2

2 2

2 2 2 2

2 2 2 22 2 2 2

2 ( ) 2 ( ) 4

3 02 2 ( ) 2 2 2 ( )

Equal when or

x

y

x y x x y x x y xyu u

x y x y x yy x x

xy x x y y xy x x yv v

x y x y x y

+ = = =

+ + + = =

+ = = =+ + +

( ) ( )

( ) ( )

2 2 2 2 2 2 2

2 2 2 22 2 2 2

2 2

2 2 2 2

2 2 2 22 2 2 2

2 ( ) 2 ( ) 4

3 02 2 ( ) 2 2 2 ( )

Equal when or

Both Cauchy-Riemann equations cannot be satisfied

y

x

x y y x y y x y yxu u

x y x y x yx y y

xy y x y x xy y x yv v

x y x y x y

+ = = =

+ + + = =

+ = = =+ + +

Note: it is easier to use polar C-R equations: ur = v/r, vr=u/r (here u=cos(2), v=sin(2), so

ur=vr=0, while u=2sin(2) and v=2cos(2), and the polar C-R equations are not satisfied)


4/6

x1 +1

y

12=0

12=0

12=0

12=0

12=

12=+

1=0, 2=

1=2, 2=

4) (16pts) Find all branch points of tanh1z =

1 1log

2 1

z

z

+

, and find the branch cut for any branch choice

for this function (hint: this is simpler than examples we solved in class). How much does the function

jump across the branch cut(s)?

Since [ ]1 1 1

log log(1 ) log(1 )

2 1 2

zz z

z

+= +

it is clear that the branch points are 1z = and 1z =

(note though that above logarithmic identity will not hold for all branch choices)

Note thatz= is not a branch point since plugging inz=1 /tyields1 1 1/ 1 1

log log2 1 1/ 2 1

t t

t t

+ +=

It is easy to find a branch of this expression that is continuous at t=0 (take for instance the principal

branch of logarithm of (t+1)/(t1) )

We can find a branch of this multi-valued function using any of the three methods we learned in class:

METHOD 1: Chooselog( 1)

1 1 1 1

log log(1 ) log(1 ) log ( 1) log ( 1)2 1 2 2p p

i z

z

z z z z iz+

+

= + +

logp(z+1) has a brunch cut (jump by 2i) along (1, )

logp(z1) has a brunch cut (jump by 2i) along (+1,)

Subtracting the two cancels the jump everywhere except the cut

(1,+1), where the function jumps by i.

METHOD 2 (mapping) Choose

1 1 1 1

log log2 1 2 1p

x

z z

z z

+ +=

Principal branch has a cut whenx>0. Solving forz yields1

1

xz

x

=

+

. Asx varies from 0 to infinity,z

covers real interval (1,+1). Function jumps by ias this branch is crossed from above.

METHOD 3 (Independent parametrization of factors)

1 1

2 2

exp( )

1 1 21 2

2

exp( )

1 1 1 1 1log log ln , 0 , 22 1 2 1 2 2

BRANCH DEFINITION

r i

r i

rz z iz z r

+ + += = + < <

Function jumps by i as the branch is crossed from above


5/6

1 +1

y

b-cuts of log(z2+i)/2

1 +1

y

5) (16pts) Use parametrization to show that the following integral is zero over any circle around the

origin:

( )22 2 2 2

2 2 2

| | 0 0 0 0

1 1 1 1 RR R R R 1 R 0

R 2R

it it it it it it it

it

z R

z dz e ie dt e e ie dt i e dt ez e

=

+ + = + = + = + = =

Does it follow that the integrand has an anti-derivative everywhere in the domain C/{0}?

Does it follow that the integrand is analytic in this domain? Explain your answers.

No, neither of the above statements is true in this case, since the integral may well be non-zero along a

non-circular Jordan contour (one can try for instance a rectangular closed path)

6) (16pts) Without resorting to parametrization, calculate2

C

z dz

z i+along two different contours:

a) C = any contour from z = 1 to z = 1 not containing singularities of integrand

First, note that the integrand has two singularities at 1/ 2 / 2 1/ 2 / 41,2

1( ) ( )

2

i i iz i e e

= = = =

This integral does depend on the contour choice. If there are no

singularities between the contour and the real axis (see figure), we

can deform the contour to the interval of the real axis, along whichthe integral is zero sincex/(x

2 + i) is an odd function.

We can also use the Fundamental Theorem of Calculus; if the

contour does not intersect the branch cut of the anti-derivative

log(z2+i)/2 (you had to make this comment for full credit) we obtain:

1

2

2

1

1 1 1log( ) log(1 ) log(1 ) 0

2 2 2C

z dzz i i i

z i

= + = + + =

+

Optional: this is not true if there is a pole between the contour and

the real axis (see bottom figure). In this case we can close the

contour along the real axis (which does not contribute to theintegral according to above), and use the Cauchy Integral Formula:

1

2 21 2 2 1

1

2

2 1 /C

zz dz z dz ii iz z z zz i z i

= = = =

+ +

Note that this equals the jump of the anti-derivative across its branch cut (maroon dashed line)

b) C = circle of radius 2 around the origin in the positive direction

Use the Cauchy Integral Formula:

1 2

2

1 2 1 2 2 1

2 2( )( )

C C

z zz dz z dzi i

z z z z z z z zz i

= = + =

+


6/6

7) (15pts) Parametrize the integral| | 1

z

z

edz

z=

(where is a real constant) and use the Cauchy Integral

Formula to show that cos

0

cos( sin )e d

= .

By Cauchy Integral Formula we have

0

| | 1

2 2z

z

edz i e i

z

=

= =

Also using parametrizationz=eit, we obtain

{ }

2 2 2 2exp( )exp( ) cos sin cos sin

0 0 0 0

2 2 2

cos cos cos

0 0 0

cos( sin ) sin( sin ) sin( sin ) cos( sin )

ii i i i

i

eie d i e d i e d i e e d

e

i e i d e d i e d

+

= = =

= + = +

Now, the real part of this integral is zero (its an odd function), while the imaginary part has to equal

2iby the Cauchy Integral Formula, which gives us the final result

2

cos

0

cos( sin ) 2e d

=

Since the integrand is even with respect to =, we can divide the integration interval by half, yielding

the final answer of

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MATH 656 S11 Midterm Solutions