z^3=-125i

[mesty]

Hi,

I am having trouble solving z^3=-125i.

I keep getting 5(cos(theta)+isin(theta)) where theta is ((pi/2)+2kpi)/3) for k=0,1,2.

I dont think that these angles are correct.

Can somebody help me on this?

Thanks.

 

[pka]

I am having trouble solving z^3=-125i.

Let \(\displaystyle \rho = 5\exp \left( {\frac{{ - \pi i}}{6}} \right) = 5\cos \left( {\frac{{ - \pi }}{6}} \right) + 5i\sin \left( {\frac{{ - \pi }}{6}} \right)\) and \(\displaystyle \xi = \exp \left( {\frac{{2\pi i}}{3}} \right)\).

Then the roots are: \(\displaystyle \rho\cdot \xi^k,~k=0,1,2\).

 

[mesty]

why is it (-pi/6)?

I thought the angle was pi/6

Also k=0 gives 0, k=1 gives pi/2 and k=2 gives 7pi/6. Is this correct?

Thanks.

 

[mesty]

k=0 does not seem to work as it would give -pi/6 but checking to solution gives pi/2 7pi/6 and 11pi/6. Where am i going wrong?

 

[mesty]

Ok, i tried the problem again and got pi/6, 5pi/6 and 3pi/2. My result is flipped on the real axis.


z^3=-125i

r=5 and theta=(pi/2)+2kpi

z=125^(1/3)(cos((pi/2)+2kpi)+isin((pi/2)+2kpi))^(1/3)

z=5(cos(((pi/2)+2kpi)/3)+isin((pi/2)+2kpi)/3)

k=0,1 and 2 gives pi/6, 5pi/6 and 3pi/2.

What have i done wrong here?

Thanks.

 

[mesty]

I just figured it out myself I was using the wrong angle -125i is 3pi/2 not pi/2. Thanks