 ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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# Converting Complex Numbers From Rectangular Form to Trigonometric - Problem 3

Norm Prokup ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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I'm converting complex numbers from rectangular form to trigonometric form. Here is a harder example. Z equals negative root 6, minus i times root 2. Let me plot this number. I'll just plot it approximately because I don't know the exact value of root 6, or root 2. Then I'll say the negative root 6 is going to be between 2 and 3. So let's just say 2 and 1/2, -2 and 1/2. This is about -1 and 1/2.

So -2, -1/2, it will be somewhere in this direction. Now because this number z is in the third quadrant, I'm going to expect an argument that's bigger than pi, somewhere between pi and 3 pi over 2. I'll have to find r. R is the distance of the point from the origin. Let me find that first.

Remember that r is the square root of a² plus b², where a is the real part, and b is the imaginary part, the imaginary coefficient. So we have negative root 6² plus, and I should include the negative sign in b also, negative root 2². Negative root 6² is 6, negative root 2² is 2, so this is root 6 plus root 2, root 8 which simplifies to 2 root 2. That's our r value. That's the distance of the number from the origin.

Now let's find the argument. To find the argument, I need to use the fact that cosine of theta, cosine of the argument, is a over r. A is negative root 6, and r is 2 root 2. Now remember your properties of radicals. The root 2 and the root 6 cancel, leaving the root 3. So this is negative root 3 over 2.

Now the sine of theta, sine of the argument, is b over r. Remember that b is negative root 2 over 2 root 2. So this means -1/2, you have cancellation. Now what angle has a cosine of negative root 3 over 2, and a sine of -1/2? Because we're in the third quadrant, I'm going to expect that angle again to be between pi, and 3 pi over 2.

One of the things you have to be careful about, is if you're in a habit of using inverse sine, or inverse cosine, you will never get an angle in this quadrant. So you really should draw a picture, and do a reality check. I tend to not use inverse trig functions. I just remember my reference angles.

Remember that the angle with cosine of 3 pi over 2 and a sine of 1/2 is pi over 6. That's our reference angle. That means this angle is going to be pi over 6. So we're just pi over 6 past pi. That means this angle is 7 pi over 6. That's our argument.

The final trig form of this complex number is going to be z equals 2 root 2, times the cosine of 7 pi over 6 plus i sine 7 pi over 6. That's our final answer.