Unit
Polar Coordinates and Complex Numbers
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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Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
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.