When dealing with imaginary numbers, we multiply by foiling as we do with binomials. When we do this we get the expression below: 

Since we know that  we get  which gives us . 

Multiply the numerator and denominator by the numerator's complex conjugate.

Reduce/simplify.

 The first step is to distribute which gives us:

which is in standard form.

Use the FOIL method to simplify. FOIL means to mulitply the first terms together, then multiply the outer terms together, then multiply the inner terms togethers, and lastly, mulitply the last terms together.

The imaginary  is equal to:

Write the terms for .

Replace  with the appropiate values and simplify.

First remember the form of the complex number:

Where "a" is the "Real" part, and "b" is the imaginary part.

Only the real parts can be combined together, and only the imaginary parts can be combined together.

Therefore the equation becomes after distributing the negative sign:

Collect the real and imaginary terms together:

Giving the 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 

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 , .

450 ends in a "0" and is therefore a multiple of both 2 and 5. Also, since

,

450 is a multiple of 3. 

Therefore, 

.

We are looking for an element of  - that is, a multiple of 2, 3, and 5. All five choices are multiples of 5 (ending in a "5" or a "0"); 725 and 735 can be eliminated as they are not multiples of 2 (ending in a "5"). We test 720, 730, and 740 to find the multiple of 3:

720 is the correct choice.

In the above set, four of the numbers are all even .  

The only one that is not even, also making it your answer, is  since it is an odd number.

An intersection of two sets is defined as the set of elements that are members of both sets. The correct answer is the pair of sets that has no overlap.

Rationals are defined  if a and b are integers. The irrationals are defined as any real number that is not rational. By definition a number cannot be both rational and irrational.

Prime numbers are divisible only by 1 and themselves. Even numbers are defined as integer multiples of . A number common to both sets is .

Natural numbers are the counting numbers, while the integers are all the naturals and their opposites. There are many elements common to both sets, such as :

The negatives are all the integers less than zero. And remember that an even number is defined as any integer multiple of two. Again there are many elements in common including :

The naturals are the counting numbers, while the wholes are all the naturals and zero. These two sets share many elements in common such as .