Product & Quotient of Polar Complex Numbers

[generalgrievous]

You are so fun to listen to. I've never seen a teacher with such enthusiasm!

[profrobbob]

Thank you very much. I really appreciate you watching and the kind words.

[profrobbob]

@Dark27inferno You are very welcome. Thanks for watching:) Please spread the word about my new math channel.

[blahwazhere]

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[BrosephThe]

Your videos literally save my Pre calc grades, thank you thank you thank you. I learn a lot more from you than trying to learn out of my text book, and thank you for making your videos FREE. keep it up!

[profrobbob]

I have taught every level except Calculus. If I can't ever make that happen...I'll teach it to YouTube as best I can! Right now I am doing AP Statistics, PreCalculus, Honors Geometry, and Intensive Math...a remedial course. I have no planning period this year to boot!!! Last year I made Stats and Precalc vids together and this year I can barely keep up with just Geometry.

[profrobbob]

No I don't know what may be on your placement test...sorry. Can you get your hands on some review material from the school that will administer the test? I am assuming it will be a lot of concepts you learned in algebra 2. Thank you for supporting my channel, I wish I could be more helpful.

[profrobbob]

Thanks for the appreciation! I hope you will continue to watch my channel and support my efforts by spreading the word and sharing my channel with others...that's how my channel grows and what keeps these videos FREE:)

[THECRRRRRREAM]

been out of school for a while and love your videos keep doin your thing. love that your helping out the people who cant make it to school. 

[zakiazizi]

thank you Prof Rob you are a perfect teacher

[Darkinferno]

You sir are a genius. Thanks for the video! You explain it better than my teacher!

[eric]

I have been sick with bronchitis for two weeks.... Missed a week of class. Your videos are the only reason I am surviving pre-calc right now! Keep it up Professor Rob!

[georgesadler]

Professor RobBob, thank you for a solid video/lecture on the Product and Quotient of Polar Complex Numbers. Students will see this material again in Precalculus.

[stephenugochukwu]

Thank you so much professor; great lecture.

[GeoTedify]

This good and dandy, all your videos are, but I don't suppose you can tell what I typically need to know for a placement test to get into Pre-Calc? I mean I don't exactly have an entire grasp of 140 Pre-Calc videos floating in my head. I just want to know your thoughts and if you can help more than the great amount you already did.

[samschellhase]

Is it just easier to do this in polar form rather than rectangular form?

[benl]

Hi, ProfRobBob, some colleagues of mine and me are disagreeing on a solution with complex numbers.

The question: Use De Moivre's Theorem to find all the cube roots of 1

The main question is do we express the answer in rectangular form or polar form and then also for each value of k0, k1, k2 or just for one root k0?

Also if the question explicitly specifies to express the roots in form (a + bi) a, b Element R (this is rectangular, right?)

And also when the question explicitly specifies to express in polar form.

Hope that makes sense.

To try express the answers I got:
T -> theta

n = 1/3
r = 1
T = 0 degrees

k0 = r^n ( cos [(T + 2pi k) / n] + i sin [(T + 2pi k) / n])
    = 1^1/3 ( cos [(0 + 2pi 0) / 3] + i sin [(0 + 2pi 0) / 3])
    = 1 ( cos 0 + i sin 0) Polar form?
    = 1 (1 + 0) Rectangular form?

k1 = r^n ( cos [(T + 2pi k) / n] + i sin [(T + 2pi k) / n])
     = 1 ( cos [(0 + 2pi 1) / 3] + i sin [(0 + 2pi 1) / 3])
    = 1 (cos 120 + i sin 120)
    = 1 (cos 60 + i sin 60) polar form?
    = 1 (-1/2 + sqrt(3) / 2) rectangular form?

k2 = r^n ( cos [(T + 2pi k) / n] + i sin [(T + 2pi k) / n])
    = 1 ( cos [(0 + 2pi 2) / 3] + i sin [(0 + 2pi 2) / 3])
    = 1 (cos 240 + i sin 240)
    = 1 (cos 60 + i sin 60) polar form
    = 1 (-1/2 - sqrt(3)/2) rectangular form

[antonsteenberg]

Hi I had a question and not sure what to do, how do you work out the polar form when they ask that (z1)squared times (z2)cubed not sure hot to approach such a question?? Please help +profrobbob

[buthinahalashbi]

you are good teacher, but i did not find what i want ( can you make a video to show us how to solve equations in polar form please )

[kenzie]

he sounds like remy buxaplenty from fairly odd parents. no hate, just sharing my thoughts lol