All GRE Subject Test: Math Resources
Example Questions
Example Question #111 : Classifying Algebraic Functions
Simplify
In problems like this, you are expected to simplify by removing i from the denominator. To do this, multiply the numerator and denominator by the conjugate of the denominator (switch the sign between the two terms from either a plus to a minus or vice versa) over itself. The conjugate over itself equals 1 and does not change the value of the expression (any number multiplied by 1 is still that number). Multiplying by the conjugate is the only way to eliminate i since there will be no middle term when we foil.
Simplify i squared to be -1 and combine like terms
Example Question #112 : Classifying Algebraic Functions
Simplify
In problems like this, you are expected to simplify by removing i from the denominator. To do this, multiply the numerator and denominator by the conjugate of the denominator (switch the sign between the two terms from either a plus to a minus or vice versa) over itself. The conjugate over itself equals 1 and does not change the value of the expression (any number multiplied by 1 is still that number). Multiplying by the conjugate is the only way to eliminate i since there will be no middle term when we foil.
Simplify i squared to be -1 and combine like terms
Example Question #113 : Classifying Algebraic Functions
Simplify
In problems like this, you are expected to simplify by removing i from the denominator. To do this, multiply the numerator and denominator by the conjugate of the denominator (switch the sign between the two terms from either a plus to a minus or vice versa) over itself. The conjugate over itself equals 1 and does not change the value of the expression (any number multiplied by 1 is still that number). Multiplying by the conjugate is the only way to eliminate i since there will be no middle term when we foil.
Simplify i squared to be -1 and combine like terms
Each term divides by 2 so make sure to reduce all of the terms
Example Question #114 : Classifying Algebraic Functions
Simplify
In problems like this, you are expected to simplify by removing i from the denominator. To do this, multiply the numerator and denominator by the conjugate of the denominator (switch the sign between the two terms from either a plus to a minus or vice versa) over itself. The conjugate over itself equals 1 and does not change the value of the expression (any number multiplied by 1 is still that number). Multiplying by the conjugate is the only way to eliminate i since there will be no middle term when we foil.
Simplify i squared to be -1 and combine like terms
Example Question #115 : Classifying Algebraic Functions
Simplify
Undefined
In problems like this, you are expected to simplify by removing i from the denominator. To do this, multiply the numerator and denominator by the conjugate of the denominator (switch the sign between the two terms from either a plus to a minus or vice versa) over itself. The conjugate over itself equals 1 and does not change the value of the expression (any number multiplied by 1 is still that number). Multiplying by the conjugate is the only way to eliminate i since there will be no middle term when we foil.
Simplify i squared to be -1 and combine like terms
Since every term divides by 8 make sure to reduce all the terms by that greatest common factor
Example Question #111 : Classifying Algebraic Functions
Simplify
In problems like this, you are expected to simplify by removing i from the denominator. To do this, multiply the numerator and denominator by the conjugate of the denominator (switch the sign between the two terms from either a plus to a minus or vice versa) over itself. The conjugate over itself equals 1 and does not change the value of the expression (any number multiplied by 1 is still that number). Multiplying by the conjugate is the only way to eliminate i since there will be no middle term when we foil.
Simplify i squared to be -1 and then combine like terms
Example Question #161 : Imaginary Numbers
Write the complex number in polar form, where polar form expresses the result in terms of a distance from the origin on the complex plane and an angle from the positive -axis, , measured in radians.
To see what the polar form of the number is, it helps to draw it on a graph, where the horizontal axis is the imaginary part and the vertical axis the real part. This is called the complex plane.
To find the angle , we can find its supplementary angle and subtract it from radians, so .
Using trigonometric ratios, and .
Then .
To find the distance , we need to find the distance from the origin to the point . Using the Pythagorean Theorem to find the hypotenuse , or .