Cube Roots of a Complex Function Am I doing something wrong?

[kawsar]

1. Find the cube roots of the complex number 8+8i and plot them on an Argand
diagram

Thats the problem, I've had a go at the solution and came up with 3 solutions using the [tex]\sqrt[n]{r}[/tex]*(cos([tex]\frac{\theta+2\pi*k}{n}[/tex])+isin([tex]\frac{\theta+2\pi*k}{n}[/tex])), but the answers (roots) I get, I can't plot it on an Argand Diagram as I cannot simplify them to an a+bi format.

I'm using n=3, r=8[tex]\sqrt{2}[/tex] and [tex]\theta[/tex]=45 degrees.

Am I doing something wrong somewhere?

Thanks

First Post :D

 

[Mark44]

One of the roots should have a magnitude of r = 2*2^(1/6), and an argument (angle) of 15 degrees. That should agree with your De Moivre formula for k = 0. For the other two roots, use k = 1 and k = 2.

 

[kawsar]

Yep, that's what I get as well, but how would I plot that root on an Argand Diagram? I could work out the a and b (to make it into the format a+bi) using a calculator but I'm not allowed to use one you see.

Any ways I could do that without the use of a calculator etc.?

Thanks!

 

[Mark44]

You can't use a calculator at all? I can see not allowing a graphing calculator, but can't you use one to get an approximation for 2*2^(1/6) and for the sine and cosine of 15 degrees?

Well, the angle is 15 degrees, so you could use some half-angle formulas to get sin(15 deg) and cos(15 deg). The modulus is a little over 2 (~2.245), so draw a point on a circle of radius 2.245, and then make a ray 15 degrees up from the positive x axis. Where the ray crosses the circle is one of your cube roots.

 

[kawsar]

Here is what it says on the past exam paper it came from: Calculators are NOT permitted in this examination.

So, no calculators of any kind allowed by the looks of it. I might have to email one of the lecturers to see what they say about this. Seems a tough one having to work out the modulus (approx).

The question does not say approximate but it could be that when we make the plots we just plot them approximately (2 and a bit) as I'm sure they won't be able to tell what 2.245 and a little over 2 is.

Thanks for the replies.

 

[Mark44]

Yeah, that sounds like a good strategy of just making r out a bit over 2. And if you're halfway careful you can get a reasonable set of values for x and y just by eyeball.

Actually, I think it's a good idea for schools to not allow calculators for some problems.

 

[kawsar]

I thought it was the standard that Universities don't allow Calculators at all... Makes you think more (which is better) but some questions become a tad longer which doesn't help when you're running out of time...!