## SAT 2016

## CHAPTER 10

## THE SAT MATH: ADDITIONAL TOPICS

### Skill 3: Understanding Complex Numbers

**Lesson 11: Understanding the imaginary number i and the complex plane**

**Imaginary Numbers**

The **imaginary number** *i* is defined as the principal square root of –1.

The square root of a negative number is not on the real number line, because the square of a real number cannot be negative. Therefore, the square roots of negative numbers must reside on their own number line, which we call the **“imaginary axis”**, which is perpendicular to the real axis, intersecting it at the origin. The plane defined by the real axis and the imaginary axis is called the **complex plane**.

Given that , which of the following is equal to ?

A) *i*

B) −*i*

C) 1

D) −1

(*Medium-hard*) To answer this question, we just need to know the basic exponent rules and the definition of *i*.

Original expression:

Factor:

Substitute *i*^{2} = –1

Multiply by *i/i*:

Substitute *i*^{2} = –1:

–(–*i*) = *i*

Therefore, the correct answer is (A).

**Lesson 12: Adding, multiplying, dividing, and simplifying complex numbers**

**Complex Numbers**

The sum of a real number and an imaginary number is called a **complex number**. All complex numbers can be expressed in the form ** a** +

**where**

*bi**a*and

*b*are real numbers and .

**Every complex number a** +

*bi*corresponds to the point (*a*,*b*) on the complex plane.**Adding Complex Numbers**

To add complex numbers, just combine “like” terms.

Original expression:

(3 – 2*i*) + (2 + 6*i*)

Regroup with Commutative and Associative Laws of Addition:

(3 + 2) + (–2*i* + 6*i*)

Simplify:

5 + 4*i*

**Multiplying Complex Numbers**

To multiply complex numbers, just FOIL and combine like terms.

Original expression:

(3 – 2*i*)(2 + 6*i*)

FOIL:

(3)(2) + (3)(6*i*) + (–2*i*)(2) + (–2*i*)(6*i*)

Simplify:

6 + 18*i* – 4*i* –12*i*^{2}

Substitute *i*^{2} = –1:

6 + 18*i* – 4*i* – 12(–1)

Combine like terms:

18 + 14*i*

**Dividing Complex Numbers**

To divide complex numbers, express the quotient as a fraction, multiply numerator and denominator by the **complex conjugate** of the denominator, and simplify. The **complex conjugate** of *a* + *bi* is *a* – *bi*.

Original expression:

Multiply numerator and

denominator

by conjugate of denominator:

FOIL:

Simplify:

Distribute division:

**Powers of i**

**The successive powers of i (i**

^{1}

**,**

*i*^{2}

**,**

*i*^{3}

**,**

*i*^{4}

**,**

*i*^{5}

**,**

*i*^{6}

**,**

*i*^{7}

**. . .) cycle counterclockwise around the unit circle in the complex plane.**

We can verify this by expanding any positive integer power of *i*.

Expression to be evaluated:

*i*^{13}

Expand:

(*i*)(*i*)(*i*)(*i*)(*i*)(*i*)(*i*)(*i*)(*i*)(*i*)(*i*)(*i*)(*i*)

Group in pairs:

(*i* × *i*)(*i* × *i*)(*i* × *i*)(*i* × *i*)(*i* × *i*)(*i* × *i*)(*i*)

Substitute *i*^{2} = –1:

(–1)(–1)(–1)(–1)(–1)(–1)(*i*)

Simplify:

*i*

This implies that *i ^{n}* = 1 if

*n*is a multiple of 4. (That is,

*i*

^{4}= 1

*, i*

^{8}= 1

*, i*

^{12}= 1, etc.) This gives us a convenient way to simplify large powers of

*i*:

**just replace the exponent with the remainder when it is divided by 4**. For instance,

*i*

^{39}=

*i*

^{3}= –

*i*, because 3 is the remainder when 39 is divided by 4.

If , where , which of the following is equal to *K*^{2}?

A) 2*i*

B) 4*i*

C) 4 + *i*

D) 4

Therefore, the correct answer is (D).

Which of the following is NOT equal to *i*^{6} – *i*^{2}?

A) *i*^{5} – *i*

B) *i*^{4}

C) 2*i*^{3} + 2*i*

D) 1 + *i*^{6}

(*Medium*) Here, we have to use our knowledge about powers of *i*. Since *i*^{6} = (*i* × *i*)(*i* × *i*)(*i* × *i*) = (–1)(–1)(–1) = –1, and *i*^{2} = –1, the given expression, *i*^{6} – *i*^{2}, is equal to (–1) – (–1) = 0. Simplifying each choice gives us

A) *i*^{5} – *i* = *i* – *i* = 0

B) *i*^{4} = 1

C) 2*i*^{3} + 2*i* = –2*i* + 2*i* = 0

D) 1 + *i*^{6} = 1 + (–1) = 0

Therefore, the correct answer is (B).

**Exercise Set 4: Complex Numbers (No Calculator)**

1

If *a* + *bi* = (1 + 2*i*)(3 – 4*i*), where *a* and *b* are constants and , what is the value of *a* + *b*?

2

If , where *a* and *b* are constants and , what is the value of *a*?

3

For what value of *b* does (*b* + *i*)^{2} = 80 + 18*i*?

4

The solutions of the equation *x*^{2} – 2*x* + 15 = 0 are and , where *a* and *b* are positive numbers. What is the value of *a* + *b*?

5

Given that , which of the following is equal to ?

A)

B)

C)

D)

6

Which of the following expressions is equal to (2 + 2*i*)^{2}?

A) 0

B) 4*i*

C) 8*i*

D) 4 – 4*i*

7

If *B*(3 + *i*) = 3 – *i*, what is the value of *B*?

A)

B)

C)

D)

8

*x*^{2} + *kx* = –6

If one of the solutions to the equation above is , what is the value of *k*?

A) –4

B) –2

C) 2

D) 4

9

If *i*^{m} = –*i*, which of the following CANNOT be the value of *m*?

A) 15

B) 18

C) 19

D) 27

**EXERCISE SET 4: COMPLEX NUMBERS ANSWER KEY**

**No Calculator**

__1__. **13**

(1 + 2*i*)(3 – 4*i*)

FOIL:

(1)(3) + (1)(–4*i*) + (2*i*)(3) + (2*i*)(–4*i*)

Simplify:

3 – 4*i* + 6*i* – 8*i*^{2}

Substitute *i*^{2} = –1:

3 – 4*i* + 6*i* –8(–1)

Combine like terms:

11 + 2*i*

Therefore, *a* = 11 and *b* = 2, so *a* + *b* = 13.

__2__. **7/5 or 1.4**

Multiply conjugate:

FOIL:

Substitute *i*^{2} = –1:

Combine like terms:

Distribute division:

__3__. **9**

(*b* + *i*)^{2}

FOIL:

(*b* + *i*)(*b* + *i*) = *b*^{2} + *bi* + *bi* + *i*^{2}

Substitute *i*^{2} = –1:

*b*^{2} + *bi* + *bi* – 1

Combine like terms:

(*b*^{2} – 1) + 2*bi*

Since this must equal 80 + 18*i*, we can find *b* by solving either *b*^{2} – 1 = 80 or 2*b* = 18. The solution to both equations is *b* = 9.

__4__. **15** The equation we are given is a quadratic equation in which *a* = 1, *b* = –2, and *c* = 15. Therefore, we can use the quadratic formula:

Quadratic Formula:

Substitute:

Simplify:

Simplify:

Distribute division:

Therefore, *a* = 1 and *b* = 14, so *a* + *b* = 15.

__5__. **B**

FOIL:

Substitute *i*^{2} = –1:

Simplify:

Multiply by *i/i*:

Substitute *i*^{2} = -1:

__6__. **C**

(2 + 2*i*)^{2}

FOIL:

(2 + 2*i*)(2 + 2*i*) = 4 + 4*i* + 4*i* + 4*i*^{2}

Substitute *i*^{2} = –1:

4 + 8*i* – 4 = 8*i*

__7__. **D**

*B*(3 + *i*) = 3 – *i*

Divide by 3 + *i*:

FOIL:

Substitute *i*^{2} = –1:

Simplify:

Distribute division:

__8__. **B**

*x*^{2} + *kx* = –6

Add 6:

*x*^{2} + *kx* + 6 = 0

Substitute :

FOIL:

Simplify:

Distribute:

Collect terms:

Therefore, both 2 + *k* = 0 and . Solving either equation gives *k* = –2.

__9__. **B** As we discussed in Lesson 10, the powers of *i* are “cyclical,” and *i*^{m} = –*i* if and only if *m* is 3 more than a multiple of 4. The only number among the choices that is not 3 more than a multiple of 4 is (B) 18.