G
Guest
Guest
Can you help me with this one?
Find the three cube roots of -64?
Please can you get me started?
Find the three cube roots of -64?
Please can you get me started?
pka
Elite Member
- Joined
- Jan 29, 2005
- Messages
- 11,976
Change –64 into polar form.
\(\displaystyle \L
- 64 = 64\left[ {\cos (\pi ) + i\sin (\pi )} \right]\)
The three cube roots are:
\(\displaystyle \L
r^{1/3} \left[ {\cos \left( {\frac{{\pi + 2k\pi }}{3}} \right) + i\sin \left( {\frac{{\pi + 2k\pi }}{3}} \right)} \right]\quad ,k = 0,1,2\)
\(\displaystyle \L
- 64 = 64\left[ {\cos (\pi ) + i\sin (\pi )} \right]\)
The three cube roots are:
\(\displaystyle \L
r^{1/3} \left[ {\cos \left( {\frac{{\pi + 2k\pi }}{3}} \right) + i\sin \left( {\frac{{\pi + 2k\pi }}{3}} \right)} \right]\quad ,k = 0,1,2\)
skeeter
Elite Member
- Joined
- Dec 15, 2005
- Messages
- 3,216
another method ...
\(\displaystyle x^3 + 64 = 0\)
\(\displaystyle x^3 + 4^3 = 0\)
\(\displaystyle (x+4)(x^2 - 4x + 16) = 0\)
\(\displaystyle x+4 = 0\)
\(\displaystyle x = -4\)
\(\displaystyle x^2 - 4x + 16 = 0\)
\(\displaystyle x = 2 \pm 2\sqrt{3}i\)
\(\displaystyle x^3 + 64 = 0\)
\(\displaystyle x^3 + 4^3 = 0\)
\(\displaystyle (x+4)(x^2 - 4x + 16) = 0\)
\(\displaystyle x+4 = 0\)
\(\displaystyle x = -4\)
\(\displaystyle x^2 - 4x + 16 = 0\)
\(\displaystyle x = 2 \pm 2\sqrt{3}i\)
G
Guest
Guest
pka said:
I really appreciate your reply. I need to work this out using the polar form. Sorry if this seems a stupid question but how do I work out what pi is using this method?
