All Algebra II Resources
Example Question #153 : Imaginary Numbers
are real numbers.
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 #154 : Imaginary Numbers
If and are real numbers, and , what is if ?
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:
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 #155 : Imaginary Numbers
Solve for if .
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 #156 : Imaginary Numbers
Evaluate and simplify .
None of the other answers.
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 #157 : Imaginary Numbers
Solve for :
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 #4 : Imaginary Numbers & Complex Functions
Solve for and :
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 #158 : Imaginary Numbers
All real numbers
Subtract from both side:
Which is never true, so there is no solution.