Complex Numbers/Polar Form - Trigonometry

To solve this problem, we must first identify the conjugate of this complex number. The conjugate keeps the real portion of the number the same, but changes the sign of the imaginary part of the number. Therefore the conjugate of is . Now, we need to multiply these together using distribution, combining like terms, and substituting .

To solve this problem, we must understand what a complex conjugate is and how it relates to a complex number. The conjugate of a number is . Therefore the conjugate of is .

To add complex numbers, we must combine like terms: real with real, and imaginary with imaginary.

In order to solve this problem, we must combine real numbers with real numbers and imaginary numbers with imaginary numbers. Be careful to distribute the subtraction sign to all terms in the second set of parentheses.

To solve this problem, make sure you set it up to multiply the entire parentheses by itself (a common mistake it to try to simply distribute the exponent 2 to each of the terms in the parentheses.)

(recall that )

Please note that while the answer choice is not incorrect, it is not fully simplified and therefore not the correct choice.

While these terms may not look like they follow the typical format of , don't let them fool you! We can read 5 as and we can read 3i as . Now recalling that the complex conjugate of is , we can see that the complex conjugate of is just and the complex conjugate of is

To perform division on complex numbers, multiple both the numerator and the denominator of the fraction by the complex conjugate of the denominator. This looks like:

To represent complex numbers graphically, we treat the x-axis as the "axis of reals" and the y-axis as the "axis of imaginaries." To plot , we want to move 6 units on the x-axis and -3 units on the y-axis. We can plot the point P to represent , but we can also represent it by drawing a vector from the origin to point P. Both representations are in the diagram below.

We can take any complex number and graph it as a vector, measuring units in the x direction and units in the y direction. Therefore . Likewise, . Then, we can add these two vectors together, summing their real parts and their imaginary parts to create their resultant vector . Therefore the correct answer is .

The above image is a graphic representation of subtraction of complex numbers (which are represented by vectors and . We can take any complex number and graph it as a vector, measuring units in the x direction and units in the y direction. Therefore . Likewise, .

To help us visualize subtraction, instead of thinking about taking , we should instead visualize . The below figure shows with a dotted line. Visually, the resultant vector lies in between the vectors and . Algebraically, we get or . Either way you think about it, the resulting vector is

We can use DeMoivre's formula which states:

Now plugging in our values of and we get the desired result.

First, convert this complex number to polar form.

Since the point has a positive real part and a negative imaginary part, it is located in quadrant IV, so the angle is .

This gives us

To evaluate, use DeMoivre's Theorem:

DeMoivre's Theorem is

We apply it to our situation to get:

simplifying

, is coterminal with since it is an even multiple of

First convert this point to polar form:

Since this number has a negative imaginary part and a positive real part, it is in quadrant IV, so the angle is

We are evaluating

Using DeMoivre's Theorem:

DeMoivre's Theorem is

We apply it to our situation to get:

which is coterminal with since it is an odd multiplie

First, convert the complex number to polar form:

Since both the real and the imaginary parts are positive, the angle is in quadrant I, so it is

This means we're evaluating

Using DeMoivre's Theorem:

DeMoivre's Theorem is

We apply it to our situation to get.

First, evaluate . We can split this into which is equivalent to

\[We can re-write the middle exponent since is equivalent to \]

This comes to

Evaluating sine and cosine at is equivalent to evaluating them at since

This means our expression can be written as:

Begin by converting the complex number to polar form:

Next, put this in its generalized form, using k which is any integer, including zero:

Using De Moivre's theorem, a fifth root of is given by:

Assigning the values will allow us to find the following roots. In general, use the values .

These are the fifth roots of .

Begin by converting the complex number to polar form:

Next, put this in its generalized form, using k which is any integer, including zero:

Using De Moivre's theorem, a fifth root of 1 is given by:

Assigning the values will allow us to find the following roots. In general, use the values .

These are the cube roots of 1.

Begin by converting the complex number to polar form:

Next, put this in its generalized form, using k which is any integer, including zero:

Using De Moivre's theorem, a fifth root of is given by:

Assigning the values will allow us to find the following roots. In general, use the values .

These are the fifth roots of .

Given the polar coordinates , the -coordinate is . We can find this coordinate by substituting :

Likewise, given the polar coordinates , the -coordinate is . We can find this coordinate by substituting :

Therefore the rectangular coordinates of the point are .

To convert this number to rectangular form, first think about what and are equal to. Because , we can use a 30-60-90o reference triangle in the 3rd quadrant to determine these values.

Now plug these in and continue solving: