Roots of complex numbers

11 - 4 Roots of Complex Numbers

Try the quiz at the bottom of the page! go to quiz

We already know from Algebra II that every equation of power 2 has 2 solutions and every equation of power 3 has 3 solutions etc, etc ad nauseum. But up until now, we have only one solution for the equation x³ = 8. The only one you could find was x =2. Where are the other two? We know they exist, but how do you find them. If we extend De Moivre's to find roots, suprise!! We can find the missing roots!! How you might ask? Good question! I guess I will show you! Hang on to your hat and off we go!

Important definition ( This means pay attention!) The n nth roots of z = r cis 0 are: z^1/n = r^1/n cis ( 0/n + k ^. 360^o /n) for k = 0, 1, 2, 3, ... n-1

Say what? What are you trying to tell me? Here's the deal! Say you want to find the 3 cubed roots of 8. That is x³ = 8. Change 8 into r cis 0. You remember that right! 8 is on the x-axis so r cis 0 = 8 cis 0^o. Since we are working with the third power, k will equal 0, 1, and 2. Look at the formula! The highest k goes to is n-1. Tada! Now plug and chug! One at a time. When k = 0, z^1/3 = 8^1/3 cis ( 0/3 + 0 ^. 360/3) = 2 cis( 0 + 0) = 2cis 0 = 2cos 0 + 2i sin 0 = 2 Voila! When k=1, z^1/3 = 8^1/3 cis ( 0/3 + 1^. 360/3) = 2 cis ( 0+120) = 2cis 120 = __ 2cos120 + 2i sin 120 = -1 + i \/ 3 . How about that! An imaginary root. Imagine that( pun intended) When k=2, z^1/3 = 8^1/3 cis ( 0/3 + 2 ^. 360/3) = 2 cis(0 + 240) = 2cis 240 = __ 2cos240 + 2i sin 240 = -1 - i\/ 3 . Another imaginary number. Hey, I seem to recall something about these guys showing up in pairs? Do you?

So how do you go about proving what you really have are the roots of 8. How do you check? Any brilliant ideas? That's right, you multiply. What? You heard me, Multiply. Proof: 2 x 2 x 2 = 8 ( boy that was tough!) __ __ __ Proof: (-1 -i\/ 3)(-1 -i \/ 3)(-1 -i \/3 ) You guessed it FOIL! __ __ (-2 + 2i \/ 3)( -1 - i \/ 3) = 8 (Ripleys believe it or not!) __ __ __ Proof: (-1 + i \/ 3)(-1 + i \/ 3)( -1 + i \/ 3) = __ __ (-2 - 2i \/ 3)( -1 - i \/ 3) = 8 ( believe it if you must)

Basically, if you can do a couple of these monsters, you are doing pretty well. Thus ends chapter 11 and our glorious trek through the cyberspace of trigonometry. But stay tuned, it's not quite over. Up next is what you have waited for with baited breath! The sample test. To go onto the sample test hit : Sample test To back up and review before disaster strikes hit: Previous page

Current quizaroo # 11

a) (4, 45^o); b) (3, 60^o); c) (5, 45^o); d) (5, 53.1^o); e) (4, 53.1^o)

: b) limacon; c) 8-leaved rose; d) cardioid; e) lemniscate
3) Write (-3 + 4i) in polar form.: a) 5cis 53.1^o; b) 5cis 126.9^o; c) 4cis 53.1^o; d) 3cis 126.9^o; e) 5cis 233.1^o


4) Give the polar form of (2cis 45^o)³: a) 8cis 135^o; b) 8cis 45^o; c) 2cis 45^o; d) 2cis 135^o; e) 6cis 45^o

5) Find the three cubed roots of 27.

a) 3, 3i, -3i

b) 3, (-3 + 3i \/ 3 )/2

c) 3, (-1 + i \/ 3 )

d) 3, -3, \/ 3

e) 3, (-3 + 3i \/ 3 )

click here for answers!!