Solucionario Glyn James

1 Functions of a Complex Variable Exercises 1.2.2
1(a) If | z − 2 + j |=| z − j + 3 | so that | x + jy − 2 + j |=| x + jy − j + 3 | or

(x − 2)2 + (y + 1)2 = (x + 3)2 + (y − 1)2 x2 − 4x + 4 + y 2 + 2y + 1 = x2 + 6x + 9 + y 2 − 2y + 1

Cancelling the squared terms and tidying up y= 5 5 x+ 2 4

1(b)

z + z ∗ + 4j(z − z ∗ ) = 6

Using z + z ∗ = 2x, z − z ∗ = 2jy gives 2x + 4j2jy = 6 3 1 y= x− 4 4

2

The straight lines are | z − 1 − j | =| z − 3 + j | | z − 1 + j | =| z − 3 − j |

which, in Cartesian form, are (x − 1)2 + (y − 1)2 = (x − 3)2 + (y + 1)2 i.e. x2 − 2x + 1 + y 2 − 2y + 1 = x2 − 6x + 9 + y 2 + 2y + 1 y =x−2
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Glyn James: Advanced Modern Engineering Mathematics, Third edition

and (x − 1)2 + (y + 1)2 = (x − 3)2 + (y − 1)2i.e. x2 − 2x + 1 + y 2 + 2y + 1 = x2 − 6x + 9 + y 2 − 2y + 1 y = −x + 2 These two lines intersect at π/2 (the products of their gradients is −1 ) and y = 0, x = 2 at their intersection, i.e. z = 2 + j0 . w = jz + 4 − 3j can be written w = ejπ/2 z + 4 − 3j (since j = cos which is broken down as follows π π + j sin = ejπ/2 ) 2 2

3

z

−→ rotation

ejπ/2 z

−→ translation (0, 0) → (4, −3)ejπ/2 z + 4 − 3j = w

anticlockwise by 1 π 2 Let w = u + jv, z = x + jy

so that u + jv = j(x + jy) + 4 − 3j = jx − y + 4 − 3j i.e. u = −y + 4 v =x−3 Taking 6 times equation (2) minus equation (1) gives 6v − u = 6x + y − 22 so that, if 6x + y = 22 , we must have 6v − u = 0 so that u = 6v is the image of the line 6x + y = 22 (1) (2)

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Glyn James:Advanced Modern Engineering Mathematics, Third edition 4 Splitting the mapping w = (1 − j)z into real and imaginary parts gives

3

u + jv = (1 − j)(x + jy) = x + y + j(y − x) i.e. u = x + y v =y−x so that u + v = 2y Therefore y > 1 corresponds to u + v > 2 .

5

Since w = jz + j x = v − 1, y = −u

so that x > 0 corresponds to v > 1 .

6

Since w = jz + 1 v=x u = −y + 1

so that x > 0 ⇒v > 0 and 0 < y < 2 ⇒ −1 < u < 1 or | u |< 1 . This is illustrated below

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Glyn James: Advanced Modern Engineering Mathematics, Third edition

7 Given w = (j + parts,

√

√ 3)z + j 3 − 1 we obtain, on equating real and imaginary

√ √ √ u = x 3 − y − 1, v = x + y 3 + 3 √ √ or v 3 − u = 4y + 4 and v + u 3 = 4x

on rearranging. √ √ Thus 7(a) y = 0corresponds to v 3 − u = 4 or u = v 3 − 4 7(b) √ √ x = 0 corresponds to v + u 3 = 0 or v = −u 3 √ √ √ Since u + 1 = x 3 − y and v − 3 = x + y 3 squaring and adding gives (u + 1)2 + (v − √ √ √ 3)2 = (x 3 − y)2 + (x + y 3)2 = 4x2 + 4y 2 Thus x2 + y 2 = 1 ⇒ (u + 1)2 + (v − 7(d) √ 3)2 = 4

7(c)

√ √ Since v 3 − u = 4y + 4 and v + u 3 = 4x squaring and adding gives 4v 2 + 4u2 = 16(y + 1)2 + 16x2 oru2 + v 2 = 4(x2 + y 2 + 2y + 1)

Thus x2 + y 2 + 2y = 1 corresponds to u2 + v 2 = 8

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Glyn James: Advanced Modern Engineering Mathematics, Third edition 8(a) w = αz + β

5

Inserting z = 1+j, w = j and z = −1, w = 1+j gives the following two equations for α and β j = α(1 + j) + β from which, by subtraction, −1 = (2 + j)α so that β = 1 + j + α = or α = 1(−2 + j) 5 or 1 + j = −α + β

1 (3 + 6j) giving 5w = (−2 + j)z + 3 + 6j 5

8(b) gives

Writing w = u + jv, z = x + jy and equating real and imaginary parts 5u = −2x − y + 3 5v = x − 2y + 6

Eliminating y yields 5v − 10u = 5x or v − 2u = x Eliminating x yields 5u + 10v = −5y + 15 so that y > 0 corresponds to u + 2v < 3 or u + 2v = −y + 3

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GlynJames: Advanced Modern Engineering Mathematics, Third edition From part (b), x = v − 2u y = 3 − u − 2v

8(c)

Squaring and adding gives x2 + y 2 = (v − 2u)2 + (3 − u − 2v)2 = 5(u2 + v 2 ) − 6u − 12v + 9 | z |< 2 ⇒ x2 + y 2 < 4 so that 5(u2 + v 2 ) − 6u − 12v + 5 < 0 or (5u − 3)2 + (5v − 6)2 < 20 4 2√ 3 2 6 2 20 = = i.e. v − + v− < 5 5 5 25 5 5
2

8(d)

The fixed point(s) are given by 5z =...