# GRE Subject Test: Math : Imaginary Numbers & Complex Functions

## Example Questions

← Previous 1 3 4 5 6

### Example Question #241 : Gre Subject Test: Math

Evaluate:

Explanation:

We can set  in the cube of a binomial pattern:

### Example Question #1 : Complex Imaginary Numbers

Simplify the following product:

Explanation:

Multiply these complex numbers out in the typical way:

and recall that  by definition. Then, grouping like terms we get

### Example Question #1 : Irrational Numbers

Simplify:

Explanation:

Start by using FOIL. Which means to multiply the first terms together then the outer terms followed by the inner terms and lastly, the last terms.

Remember that , so .

Substitute in  for

### Example Question #1 : Imaginary Numbers & Complex Functions

Simplify:

Explanation:

Start by using FOIL. Which means to multiply the first terms together then the outer terms followed by the inner terms and lastly, the last terms.

Remember that , so .

Substitute in  for

### Example Question #11 : Number Theory

Simplify:

Explanation:

Start by using FOIL. Which means to multiply the first terms together then the outer terms followed by the inner terms and lastly, the last terms.

Remember that , so .

Substitute in  for

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

Solve for  and

Explanation:

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 #2 : Imaginary Roots Of Negative Numbers

Simplify:

None of the Above

Explanation:

Step 1: Split the  into .

Step 2: Recall that , so let's replace it.

We now have: .

Step 3: Simplify . To do this, we look at the number on the inside.

.

Step 4: Take the factorization of  and take out any pairs of numbers. For any pair of numbers that we find, we only take  of the numbers out.

We have a pair of , so a  is outside the radical.
We have another pair of , so one more three is put outside the radical.

We need to multiply everything that we bring outside:

Step 5: The  goes with the 9...

Step 6: The last  after taking out pairs gets put back into a square root and is written right after the

It will look something like this:

### Example Question #1 : Imaginary Numbers & Complex Functions

Explanation:

There are two ways to simplify this problem:

Method 1:

Method 2:

Explanation:
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}