Equations with Complex Numbers - Algebra 2

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

To solve

Subtract from both side:

Which is never true, so there is no solution.

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

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 , .

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

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

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

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

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 , .

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

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

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

To solve

Subtract from both side:

Which is never true, so there is no solution.

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

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

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 , .

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

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

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