AMS 10/10A, Homework 1 Solutions

For most problems, I only provide the final answers.

Problem 1:

z1

z2

= −8

5− 1

5i

(z1 + z2)3 = −2− 2i

(z1 − z̄2)−2 =

2

25+

3

50i

Problem 2:

z = − 8

13+

1

13i

Problem 3:

1. z1 = 2eπ6i, z2 =

√2e

5π4i.

2. |z1 · z2| = 2√

2, arg(z1 · z2) = 17π12

+ 2kπ, k = integer.

3. |z1/z2| =√

2, arg(z1/z2) = 11π12

+ 2kπ, k = integer.

Problem 4:

1. cosα− i sinα = cos(−α) + i sin(−α) = e−αi;

2. sinα + i cosα = cos(π2− α) + i sin(π

2− α) = e(

π2−α)i;

3. sinα− i cosα = cos(α− π2) + i sin(α− π

2) = e(α−

π2)i.

Problem 5:

1. x = 0 and x = 21/3;

2. x = 0, x = 21/3, x = 21/3e2π3i and x = 21/3e

4π3i

Problem 6:−1 + i =

√2e

3π4i

1

The solutions of z3 = −1 + i are

z = 216 e

π4i, z = 2

16 e

11π12i, z = 2

16 e

19π12i

Problem 7:i = e

π2i

The solutions of z4 = i are

z = eπ8i, z = e

5π8i, z = e

9π8i, z = e

13π8i

Problem 8:−1 + i

√3 = 2e

2π3i

The solutions of z4 = −1 + i√

3 are

z = 214 e

π6i, z = 2

14 e

2π3i, z = 2

14 e

7π6i, z = 2

14 e

5π3i

Problem 9:

1.√

3− i = 2e−1π

6i;

2. z13 = 213e−13π

6i = 213e

−1π6i = 212

√3− 212 i

3. z22 = 222e−22π

6i = 222e

2π6i = 221 + 221

√3 i

Problem 10: Let z = a+ bi.

32 = |z − (1 + i)|2 = |(a− 1) + (b− 1)i|2 = (a− 1)2 + (b− 1)2

Therefore, |z − (1 + i)| = 3 describes a circle of radius 3, centered at (1, 1).

Problem 11: Let z = reθi. √2 + i

√2 = 2e

π4i

z × (√

2 + i√

2) = (2r)e(θ+π4)i

Problem 12: For a polynomial of real coefficients, we have the decomposition

P (z) = anzn + · · ·+ a1z + a0 = an(z − ξ1) · · · (z − ξm)Q1(z)Q2(z) · · ·Qk(z)

where ξ1, ξ2, . . . , ξk are the real roots of P (z), and each Qj(z) is a quadratic polynomialcorresponding to a conjugate pair of complex roots (ζj, ζj)

Qj(z) = z2 −Re(ζj) + |ζj|2

If there is no real root, the right hand side is a product of quadratic polynomials Qj(z)’s.As a result, the degree of the right hand side is even, which contradicts with the givencondition that the polynomial is of odd degree. Therefore, a polynomial of odd degree withreal coefficients must have at least one real root.

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