### All Algebra II Resources

## Example Questions

### Example Question #1 : Equations With Complex Numbers

are real numbers.

Evaluate .

**Possible Answers:**

**Correct answer:**

For two imaginary numbers to be equal to each other, their imaginary parts must be equal. Therefore, we set, and solve for in:

### Example Question #25 : Number Theory

If and are real numbers, and , what is if ?

**Possible Answers:**

**Correct answer:**

To solve for , we must first solve the equation with the complex number for and . We therefore need to match up the real portion of the compex number with the real portions of the expression, and the imaginary portion of the complex number with the imaginary portion of the expression. We therefore obtain:

and

We can use substitution by noticing the first equation can be rewritten as and substituting it into the second equation. We can therefore solve for :

With this value, we can solve for :

Since we now have and , we can solve for :

Our final answer is therefore

### Example Question #2 : Equations With Complex Numbers

Solve for if .

**Possible Answers:**

**Correct answer:**

Go about this problem just like any other algebra problem by following your order of operations. We will first evaluate what is inside the parentheses: . At this point, we need to know the properties of which are as follows:

Therefore, and the original expression becomes

### Example Question #3 : Equations With Complex Numbers

Evaluate and simplify .

**Possible Answers:**

None of the other answers.

**Correct answer:**

The first step is to evaluate the expression. By FOILing the expression, we get:

Now we need to simplify any terms that we can by using the properties of

Therefore, the expression becomes

### Example Question #4 : Equations With Complex Numbers

Solve for :

**Possible Answers:**

**Correct answer:**

In order to solve this problem, we need to first simplify our equation. The first thing we should do is distribute the square, which gives us

Now is actually just . Therefore, this becomes

Now all we need to do is solve for in the equation:

which gives us

Finally, we get

and therefore, our solution is

### Example Question #5 : Equations With Complex Numbers

Solve for and :

**Possible Answers:**

**Correct answer:**

Remember that

So the powers of are cyclic.* *This means that when we try to figure out the value of an exponent of , we can ignore all the powers that are multiples of because they end up multiplying the end result by , and therefore do nothing.

This means that

Now, remembering the relationships of the exponents of , we can simplify this to:

Because the elements on the left and right have to correspond (no mixing and matching!), we get the relationships:

No matter how you solve it, you get the values , .

### Example Question #6 : Equations With Complex Numbers

Solve

**Possible Answers:**

All real numbers

No solution

**Correct answer:**

No solution

To solve

Subtract from both side:

Which is never true, so there is no solution.

Certified Tutor

Certified Tutor

### All Algebra II Resources

### Incompatible Browser

Please upgrade or download one of the following browsers to use Instant Tutoring: