School of Mathematical SciencesENG2092

Assignment 1 Sample Solutions2015

ENG2092 ADVANCED ENGINEERING MATHEMATICS B

Assignment 1

These sample solutions are provided to assist students in trying to see where they might have made errors intheir own work. As for most mathematical problems, there is not necessarily only one way to obtain the correctanswer, but these sample solutions are typical of what might be expected from a student who has followedthe theory and methods outlined in the ENG2092 lectures. These solutions are also intended to indicate thekind of detail expected in well-written solutions to assignments and/or examination questions in mathematics,including the use of mathematical symbols and English expression.

Note: when you complete your own assignment (or examination) solutions for this unit, it is notnecessary to restate the question. That has been done here simply to assist with reading them.

1. Calculate z = (−1 + i)/(1 + i) in two different ways. First, use the definition given in lectures. Second,convert the numbers to polar form and use the formula for division of numbers in polar form. Explain each stepin your working.

To begin, we compute directly from the definition:

−1 + i

1 + i=

(−1 + i)(1 + i)

|1 + i|2=

(−1 + i)(1− i)12 + 12

=−1 + i+ i+ 1

2=

0

2+

2

2i = i.

Next, we convert to polar form

1 + i = |1 + i|eArg(1+i) =√

2eiπ4 , −1 + i = | − 1 + i|eArg(−1+i) =

√2ei

3π4 ,

and apply the formula from lectures:

−1 + i

1 + i=

√2√2ei(

3π4 −

π4 ) = ei

π2 .

Finally, we note

ei(π2 ) := cos

(π2

)+ i sin

(π2

)= i,

as before.

2. Consider the function

f(z) =

{0 if z = 0

z2/z̄ if z 6= 0.

Determine whether f ′(0) exists. Justify your answer.

The complex function f(z) is differentiable at z = z0 if the limit

limw→0

f(z0 + w)− f(z0)

w

exists. This condition must be satisfied for all complex numbers w approaching 0.

Here z0 = 0. Let w = h where h ∈ R and h > 0. Then

limw→0

f(z0 + w)− f(z0)

w= lim

h→0

f(0 + h)− f(0)

h= lim

h→0

[1

h

(h2

h− 0

)]= lim

h→0

h2

h2= 1

since h̄ = h (because h ∈ R). Now letting w = ih where h ∈ R and h > 0, we obtain

limw→0

f(z0 + w)− f(z0)

w= lim

h→0

f(0 + ih)− f(0)

ih= lim

h→0

[1

ih

((ih)2

ih− 0

)]= lim

h→0

[1

ih

((ih)2

−ih

)]= lim

h→0

(ih)2

−(ih)2= −1.

Since these two limits do not agree, we conclude that f ′(0) does not exist.

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3. Determine all the points in the complex plane where each of the following functions are differentiable andderive a formula for the derivative at points where it exists. Explain your reasoning.

(a) f(z) = (z + 2i)/(z − 2) + (z + 2)5

(b) g(x+ iy) = (3x+ y2) + i(2xy + y)

(a) Since (z + 2i)/(z − 2) is a quotient of polynomials, it is well-defined and differentiable for all z 6= 2. Inaddition, (z + 2)5 is differentiable for all z ∈ C. Therefore

f(z) = (z + 2i)/(z − 2) + (z + 2)5

is differentiable for all z 6= 2, and using the quotient and chain rules we obtain:

f ′(z) =(z − 2)− (z + 2i)

(z − 2)2+ 5(z + 2)4

= − 2 + 2i

(z − 2)2+ 5(z + 2)4

for all z ∈ C \ {2}.

(b) Given g(x+ iy) = (3x+ y2) + i(2xy + y), let

u(x, y) = 3x+ y2, v(x, y) = 2xy + y

so that g(x+ iy) = u(x, y) + iv(x, y). Then the partial derivatives

ux = 3, uy = 2y, vx = 2y, vy = 2x+ 1

are continuous for all x+ iy ∈ C. Furthermore, since

ux − vy = 3− (2x+ 1) = 2(1− x), uy + vx = 2y + 2y = 4y,

we observe that the Cauchy-Riemann equations are satisfied if and only if x = 1 and y = 0.

Applying the theorem from lectures, we conclude that g(x+ iy) is differentiable only at the point z = 1+0i = 1.The derivative g′(1 + i0) is given by

g′(x+ i0) = (ux + ivx)|(1,0) = 3.

JH31/7/15

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