cubed roots of -64i

[tommartin]

hello,


stuck on this question with no idea how to do it
can someone please tell me what the cubed roots of -64i are?

 

[pka]

stuck on this question with no idea how to do it
can someone please tell me what the cubed roots of -64i are?

Do you know that \(\displaystyle r\text{cis}(\theta) \) is shorthand notation for \(\displaystyle r\cos(\theta)+ir\sin(\theta)~? \)

And the rule is \(\displaystyle [ r\text{cis}(\theta)]^n=[ r^n\text{cis}(n\theta)] \).

So what is \(\displaystyle \left [ 4\text{cis}\left( {\dfrac{{ - \pi }}{6}} \right) \right]^3=~? \)


Is it \(\displaystyle -64i~? \)

Now try it for \(\displaystyle \theta = \left( {\dfrac{{ - \pi }}{6} \pm \dfrac{{2\pi }}{3}} \right) \)

Look in your text for more on this topic.

 

[stapel]

can someone please tell me what the cubed roots of -64i are?

If you mean what you typed, "the cubed roots of", then you need to specify which are the roots that you're cubing. Otherwise, do you perhaps mean "the cube roots of", being the third roots, formatted as follows?

. . . . .\(\displaystyle \sqrt[3]{-64i\,}\)

Thank you!

 

[HallsofIvy]

hello,


stuck on this question with no idea how to do it
can someone please tell me what the cubed roots of -64i are?

\(\displaystyle rcis(\theta)= r(cos(\theta)+ i sin(\theta)= r e^{i\theta}\)

"-64i" has r= 64 and \(\displaystyle \theta= 3\pi/4\)

Now use \(\displaystyle (r(cos(\theta)+ i sin(\theta))^{1/3}= r^{1/3}(cos(\theta/3)+ isin(\theta/3))\)

 

[pka]

\(\displaystyle rcis(\theta)= r(cos(\theta)+ i sin(\theta)= r e^{i\theta}\)

\(\displaystyle \color{red}{-64i \text{ has }r= 64 \text{ and } \theta= 3\pi/4}\)

Come on George the \(\displaystyle \text{Arg}(-64i)=\dfrac{-\pi}{2}\). This time, who is smoking what?