Question

Find all possible roots of the complex number [125/2(1-sq(3j)]^1/3

Answer 1

no I will try and re-write the problem

Answer 2

\[\Large\bf\sf \left[\frac{125}{2}(1-\sqrt3 \mathcal i\right]^{1/3}\]

Answer 3

Like that maybe? :o

Answer 4

Darn I missed a bracket :3

Answer 5

\[[\frac{ 125 }{ 2 }(1-\sqrt{3}j)]^{\frac{ 1 }{ 3 }}\]

Answer 6

\[\Large\bf\sf =\quad \left(\frac{125}{2}\right)^{1/3}\left(\color{royalblue}{1-\sqrt3 \mathcal j}\right)^{1/3}\]Focus on the blue part for a moment.
We can rewrite these as trig functions of a specific angle.

Answer 7

Can you think of what angle that would be?

Answer 8

\[\Large\bf\sf \tan \theta\quad=\quad \frac{-\sqrt3}{1}\]

Answer 9

The real component is `positive`
the imaginary component is `negative`

so we're in the ... 4th quadrant i guess.

Answer 10

Yes Im following that ok

Answer 11

where did the \[\frac{ 5\pi }{ 3 }\] come from?

Answer 12

Oh crap I have the wrong angle... let's back up! :(
I'll try to explain that better.

Answer 13

Lemme erase some junk a minute.

Answer 14

\[\Large\bf\sf =\quad \left(\frac{125}{2}\right)^{1/3}\left(\color{royalblue}{1-\sqrt3 \mathcal j}\right)^{1/3}\]We would like to turn the blue stuff into the special values that we know from the unit circle. If we can do that, we can convert these to sine and cosine.

I don't like the thing I did with tangent, I think that was a little confusing.
Let's try this instead.
If we factor a 2 out of each term,\[\Large\bf\sf =\quad \left(\frac{125}{2}\right)^{1/3}\;2^{1/3}\left(\color{royalblue}{\frac{1}{2}-\frac{\sqrt3}{2} \mathcal j}\right)^{1/3}\]

Answer 15

Ok maybe I did have the correct angle before lol.
But uhh maybe this will make more sense... I hope at least -_-

Answer 16

Do you understand what I did with the 2?

Answer 17

Yes got that

Answer 18

So then we just need to recall at what angle:
cosine produces 1/2
sine produces -sqrt3/2