**Week 3: Complex Numbers – Day 5**

**Week 3: Complex Numbers – Day 4**

1. The non-real number i raised to what power is one? (Of course, when raised to the zeroth power it is one, but there are other powers as well.)

i^{2} = -1

i^{3} = -1 x i = -i

i^{4} = -i x i = -(-1) = 1

The fourth power of i is one. i is a fourth root of one.

2. Compute the square of:2^{-1/2}(1 + i)

The square is: (2^{-1/2})^{2}(1 + i)^{2}

= (1/2) x (1 + i) x (1 + i) = 1/2 x (1 + 2i – 1) = 1/2 x 2i = i

The square is i.

3. Compute the cube of: 2^{-1/2}(1 + i)

The cube is: i x 2^{-1/2}(1 + i)

which is: 2^{-1/2}(i – 1)

which is: 2^{-1/2}(-1 + i)

4. What power of 2^{-1/2}(1 + i) is one? In other words, 2^{-1/2}(1 + i) raised to what power is one?

Because the square is i, and i^{4} = 1, then:

((2^{-1/2}(1 + i))^{2})^{4} = 1

2^{-1/2}(1 + i) raised to the 8th power is 1.

5. Compute the square of: 2^{-1/2}(1 – i)

The square is 1/2 x (1 – i) x (1 – i) = 1/2 x (1 – 2i +(-1)) = 1/2 x (-2i)

= -i.

**Week 3: Complex Numbers – Day 3**

1. 1 / i

complex conjugate of i is -i.

= 1*(-i) / i*(-i) = -i /1 = -i

(1 divided by i).

2. (2 + 3i)/i

=(2 + 3i)*(-i) / i*(-i) = 3 – 2i

3. (1 + i) / (1 – i)

= (1 + i)*(1 + i) / (1 – i)*(1 + i) = (1 +2i – 1) / (1 – i + i + 1) = 2i/2 = i

4. i / (1 – i)

= i*(1 + i) / (1 – i)*(1 + i) = (i – 1) / (1 – i + i + 1) = -1/2 + i/2

5. (3 – 2i) / (3 + 2i)

= (3 – 2i)*(3 – 2i) / (3 + 2i)*(3 – 2i) = (9 – 6i – 4) / (3 + 6i – 6i + 4) = (5 – 6i) / 7 = 5/7 – 6i/7

**Week 3: Complex Numbers – Day 2**

Complex Solutions

1. What are the solutions of x^{2} + 9 = 0?

x^{2} = -9

Take square roots of both sides:

x = 3i or -3i

2. What are the solutions of (x + 1)^{2} + 4 = 0?

(x + 1)^{2} = -4

Take square roots of both sides:

x + 1 = 2i or -2i

Solutions are -1 + 2i or -1 – 2i

3. What are the solutions of x^{2} – 2x + 10 = 0?

Completing the square: x^{2} – 2x + 1 – 1 + 10 = 0

(x – 1)^{2} + 9 = 0

(x – 1)^{2} = -9

Take square roots of both sides:

x – 1 = 3i or -3i

Solutions: 1 + 3i or 1 – 3i

4. What are the solutions of x^{3} + 25x = 0?

Factoring: x(x^{2} + 25) = 0

Solutions are x = 0 and x^{2} + 25 = 0.

Solving the second equation:

x^{2} = -25

Take square roots of both sides:

x = 5i or -5i

Three solutions are: 0, 5i, and -5i

5. What are the solutions of (x – 1)^{3} + 36(x – 1) = 0?

Factoring: (x – 1)((x-1)^{2} + 36) = 0

Solutions: x – 1 = 0 (x = 1), and solutions of (x-1)^{2} + 36 = 0

Solving the second equation:

(x – 1)^{2} = -36

Take square roots of both sides:

(x – 1) = 6i or -6i

x = 1 + 6i or 1 – 6i

**Week 3: Complex Numbers – Day 1**

Simplify these expressions. The final result should be a number of the form A + Bi.

1. (-1 + i) + (3 + 2i)

= -1 + 3 + i + 2i = 2 + 3i

2. (7 + 2i) + 10

= 17 + 2i

3. (2 + 2i) x 3i

= 6i + 6x(-1) = -6 + 6i

4. (5 + 3i) x (5 – 3i)

= 25 + 15i – 15i + 9 = 34 = 34 + 0i

5. (4 + i) x (1 + 6i)

= 4 + 24i + i – 6 = -2 + 25i