MATH 3007A Test 1 Solutions

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MATH 3007A
Test 1 Solutions
September 30, 2011
1. Let z = 4+6i
.Expresszas a + bi and find z, Re(z), Im(z) and|z|.
1 − i
Solution:
z = 4+6i1+i
−2+10i
= = −1+5i, z = −1 − 5i, Re(z) =−1, Im(z) =5and
1 − i 1+i 2
|z| = √ 26.
(2 + 3i)4
2. Let z =
(6 − i) 3 .Findzand |z|2 . (You need not express z or z as a + bi.)
Solution:
(2 − 3i)4
z =
(6 + i) 3 , |z|2 (2 + 3i)4
= zz =
(6 − i) 3
(2 − 3i) 4 134
= .
(6 + i) 3 73 3. Let z = 3
2 − 3√3 2 i.
(a) Find all polar forms of z.
Solution:
|z| =3 ⇒ z =3
1
2 −
√
3
2 i
πi
−
=3e 3 =3e 5πi
3 =3e 5πi
3 +2πik ,kɛZ.
(b) Find arg(z) andlog(z) ifarg(z) ɛ [0, 2π].
Solution:
arg(z) = 5π
, log(z) =ln(3)+5πi
3 3 .
(c) Find arg(z) andlog(z) ifarg(z) ɛ [−π, π].
Solution:
arg(z) =− π
πi
, log(z) =ln(3)−
3 3 .
2πi
− 4. Let z =2e 3 .Findarg(z) andlog(z) if
(a) arg(z) ɛ [0, 2π].
Solution:
arg(z) = 4π
, log(z) =ln(2)+4πi
3 3 .
(b) arg(z) ɛ [−π, π].
Solution:
arg(z) =− 2π
2πi
, log(z) =ln(2)−
3 3 .
5. Find the cube roots of −i. Express them in the form a + bi.
Solution:
πi
− −i = e 2 +2πik ,k=0, 1, 2.Thecuberootsare
πi
(− e 2 +2πik)/3 = e πi
6 (−1+4k) √ ,k=0, 1, 2, i.e.,
πi
− 3 1 πi
e 6 = − i, e 2 = i and e
2 2 4πi
3 = − 1
2 −
√
3
2 i.

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6. Find the fifth roots of 1 − i. You may leave them in polar form.
Solution:
|1 − i| = √ 2 ⇒ 1 − i = √
1√2
2 − i 1
√ =
2
√ πi
−
2e 4 = √ πi
−
2e 4 +2πik ,k=0, 1, 2, 3, 4.
Thefifthrootsare21/10 πi
(− e 4 +2πik)/5 =21/10e πi
20 (−1+8k) , k =0, 1, 2, 3, 4, i.e.,
πi
− 20 , 21/10e 7πi
20 , 21/10e 3πi
4 , 21/10e 23πi
20 and 21/10e 31πi
2 1/10 e
20 .
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