**Complex Numbers**

Addition and Subtraction

Addition and Subtraction

Yesterday we began discussing complex numbers, and I gave you a problem to try involving raising *i* to a power. You can see that post here: Complex Numbers – Examples and Powers of *i*

Today I will go over how to add and subtract complex numbers. But first I will provide a solution to yesterday’s problem. Here is the problem one more time, followed by a solution:

**Example: **Compute *i*^{53}.

**Solution: **When we divide 53 be 4 we get a remainder of 1. So *i*^{53 }= *i*^{1 }= **i**.

**Notes: **(1) For a review of how to compute remainders, see the following post: Remainders

(2) This computation can also be done quickly in your calculator, but be careful. Your calculator may sometimes “disguise” the number 0 with a tiny number in scientific notation. For example, when we type i^ 53 ENTER into out TI-84, we get an output of –3E–13 + *i*. The expression –3E–13 represents a tiny number in scientific notation which is essentially 0. So this should be read as 0 + i = i.

**Addition and Subtraction**

We add two complex numbers simply by adding their real parts, and then adding their imaginary parts.

(*a* + *bi*) + (*c* + *di*) = (*a* + *c*) + (*b* + *d*)*i*

**Example: **The sum of the complex numbers 2 – 3*i* and –5 + 6*i* is:

(2 – 3i) + (–5 + 6*i*) = (2 – 5) + (–3 + 6)*i* = **–3 + 3 i**

Subtraction is similar.

(*a* + *bi*) – (*c* + *di*) = (*a* – *c*) + (*b* – *d*)*i*

**Example 5: **When we subtract 2 – 3*i* from –5 + 6i we get what complex number?

I will provide a solution to this question tomorrow. Meanwhile, please post your attempted solutions in the comments below.

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**Complex Numbers**

A **complex number** can be written in the form *a* + *b**i* where *a *and *b *are real numbers and *i*^{2} = –1.

**Examples: **The following are examples of complex numbers.

2 + 3*i*

3/2 + (–2*i*) = 3/2 – 2*i *

* *–*π* + 2.6*i*

0 + 5*i *= 5*i * This is called a **pure imaginary** number.

17 + 0*i *= 17 * * This is called a **real number.**

0 + 0*i* = 0* *This is **zero**.

**Powers of i**

By definition, when we raise any number to the 0 power, we get 1. So in particular, we have *i*^{0} = 1.

Similarly, when we raise any number to the power of 1, we just get that number. So *i*^{1} = *i*.

By the definition of *i*, we have *i*^{2} = –1.

*i*^{3} = (*i*^{2})(*i*) = (–1)*i* = –*i.*

*i*^{4} = (*i*^{2})(*i*^{2}) = (–1)(–1) = 1*.*

*i*^{5} = (*i*^{4})(*i*) = (1)(*i)* = *i.*

⋯

Notice that the pattern begins to repeat.

Starting with *i*^{0} = 1, we have

*i*^{0} = 1 *i*^{1} = *i* *i*^{2} = –1 *i*^{3} = –*i*

*i*^{4} = 1 *i*^{5} = *i* *i*^{6} = –1 *i*^{7} = –*i*

*i*^{8} = 1 *i*^{9} = *i* *i*^{10} = –1 *i*^{11} = –*i*

⋯

In other words, when we raise *i* to a nonnegative integer, there are only four possible answers:

1, *i*, –1, or –*i*

To decide which of these values is correct, we can find the remainder upon dividing the exponent by 4.

**Example: **Compute *i*^{53}.

I will provide a solution to this question tomorrow. Meanwhile, please post your attempted solutions in the comments below.

For those of you that prefer videos…

Check out the Get 800 collection of test prep books to learn how to apply this information to standardized test questions.

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**Inverse Variation Question 4 with Solution**

Last week, I went over inverse variation, and I gave you four inverse variation problems to try on your own. You can see that post here: Inverse Variation

Today I would like to give a solution to the last of those four problems. You can see a solution to the first three problems here: Q1 Q2 Q3

**Example: **Suppose that *z* varies directly as *x*^{2} and inversely as *y*^{3}. If *z *= 9 when *x *= 3 and *y *= 2, what is *y* when *z *= 4.5 and *x *= 6 ?

Try to solve the problem yourself before checking the solution below.

**Solution: **We are given that *z* = *k**x*^{2}/*y*^{3} for some constant *k*. Since z = 9 when *x* = 3 and *y* = 2, we have 9 = k(3)^{2}/2^{3} = 9*k*/8. So *k* = 8, and *z* = 8*x*^{2}/*y*^{3} . We now substitute *z* = 4.5 and *x* = 6 to get 4.5 = 8(6)^{2}/*y*^{3} . So *y*^{3} = 8(36)/4.5 = 64, and therefore *y* = 4, choice D.

**More Problems with Explanations**

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**Inverse Variation Question 3 with Solution**

Last week, I went over inverse variation, and I gave you four inverse variation problems to try on your own. You can see that post here: Inverse Variation

Today I would like to give a solution to the third of those four problems. You can see a solution to the first two problems here: Q1 Q2

**Example: **If *x* ≠ 0 and *x* is directly proportional to *y*, which of the following is inversely proportional to 1/*y*^{2} ?

A) *x*^{2}

B) *x*

C) 1/*x*

D) 1/*x*^{2}

Try to solve the problem yourself before checking the solution below.

**Solution: **Since *x* is directly proportional to *y*, *x* = *ky* for some constant *k*.

Taking the reciprocal of each side of this equation gives us 1/*x* = 1/*ky*.

Cross multiplying yields *ky* = *x*.

So *k* = *x*(1/*y*). Squaring each side of this last equation gives *k*^{2 }= *x*^{2}(1/*y*^{2}).

Since *k*^{2} is a constant, *x*^{2} is inversely proportional to 1/*y*^{2}. So the answer is choice A.

**More Problems with Explanations**

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**Inverse Variation Question 2 with Solution**

Last week, I went over inverse variation, and I gave you four inverse variation problems to try on your own. You can see that post here: Inverse Variation

Today I would like to give a solution to the second of those four problems. You can see a solution to the first problem here: Inverse Variation Q1

**Example: **If *x* varies inversely as *y*^{2}, and *x* is 3 when *y* is 5, then what is *x* when *y* is 3 ?

Try to solve the problem yourself before checking the solution below.

**Solution: **We are given that *y* = 5 when *x* = 3. Since *x* varies inversely as *y*^{2}, we have that *xy*^{2 }is a constant. So we get the following equation:* *(3)(25) = *x*(9). So 75 = 9*x*, and therefore *x* = 75/9 = **25/3**.

**More Problems with Explanations**

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**Inverse Variation Question 1 with Solutions**

Today I would like to give a solution to the first of those four problems. I will give solutions to the other three problems throughout this week.

**Example: **If *y* = *k*/*x* and *y* = 9 when *x* = 14, then what is *y* when *x* = 6 ?

Try to solve the problem yourself before checking the solutions below.

**Solutions: **

**(1) **We are given that *y* = 9 when *x* = 14, so that 9 = *k*/14, or *k* = 126. Thus, *y* = 126/*x*. When *x* = 6, we have *y* = 126/6 = **21**.

**(2) **Since *y* varies directly as *x*, *xy** *is a constant. So, we get the following equation: (14)(9) = 6*y*. So, 126 = 6*y*, and *y* = 126/6 = **21**.

**(3) **(14)(9)/6 = **21**.

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Inverse Variation

Inverse Variation

Last week, I gave you several posts on direct variation. You can see those posts by clicking the following links: Direct Variation Q1 Q2 Q3

Today I would like to talk about **inverse variation**.

The following are all equivalent ways of saying the same thing:

(1) *y* varies inversely as *x. *

(2) *y* is inversely proportional to *x.*

(3) *y *= *k*/*x *for some constant *k.*

(4) *xy *is constant.

The following is a consequence of (1), (2), (3), or (4)

(5) the graph of *y *= *f*(*x*) is a hyperbola

**Note:** (5) is not equivalent to (1), (2), (3), or (4).

**Example:** In the equation *y* = 12/*x*, *y* varies inversely as *x*. Here is a partial table of values for this equation.

Note that we can tell that this table represents an inverse relationship between *x* and *y* because (1)(12) = (2)(6) = (3)(4) = (4)(3) = 12. Here the **constant of variation** is 12.

Here is a graph of the equation. On the left, you can see the full graph. On the right, we have a close-up in the first quadrant.

The various equivalent definitions of direct variation lead to several different ways to solve problems.

**Example: **If *y* = *k*/*x *and *y* = 8 when *x* = 3, then what is *y* when *x* = 6 ?

Try to solve the problem yourself before checking the solutions below.

**Solutions: **

**(1) **We are given that *y* = 8 when *x* = 3, so that 8 = *k*/3, or *k* = 24. Thus, *y* = 24/x. When *x* = 6, we have *y* = 24/6 = **4**.

**(2) **Since *y* varies directly as *x*, *xy** *is a constant. So, we get the following equation: (3)(8) = 6y. So, 24 = 6*y*, and *y* = 24/6 = **4**.

**(3) **(8)(3)/6 = **4**.

Here are a few more problems for you to try. I will provide solutions to these over the next few days.

1. If *y* = *k*/*x *and *y* = 9 when *x* = 14, then what is *y* when *x* = 6?

2. If *x* varies inversely as *y*^{2}, and *x* is 3 when *y* is 5, then what is *x* when *y* is 3

3. If *x* ≠ 0 and *x* is directly proportional to *y*, which of the following is inversely proportional to 1/*y*^{2} ?

A) *x*^{2}

B) *x*

C) 1/*x*

D) 1/*x*^{2}

4. Suppose that *z* varies directly as *x*^{2} and inversely as *y*^{3}. If *z *= 9 when *x *= 3 and *y *= 2, what is *y* when *z *= 4.5 and *x *= 6 ?

**More Problems with Explanations**

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