Complex Numbers help

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kmalik001

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So could someone give me a headstart on the question please. I understand that v and its conjugate are roots of z but I just don't know what the other 4 roots are going to be since I guess there are 6 roots in total becuase z is to the power of 6.


Determine the complex constant c such that v is a root of
z^6 − c = 0
Plot all roots in the complex plane.
v= √3 − i
 
kmalik001 said:
So could someone give me a headstart on the question please. I understand that v and its conjugate are roots of z but I just don't know what the other 4 roots are going to be since I guess there are 6 roots in total becuase z is to the power of 6.


Determine the complex constant c such that v is a root of
z^6 − c = 0
Plot all roots in the complex plane.
v= √3 − i
Click to expand...

v = r * eiΘ = 2 * e-i(π/6) ......................... mistake corrected

and continue....
 
Last edited by a moderator:
well you just converted into re(j*theta) form. What should I do with that? Could you elaborate the process please.
 
kmalik001 said:
So could someone give me a headstart on the question please. I understand that v and its conjugate are roots of z but I just don't know what the other 4 roots are going to be since I guess there are 6 roots in total becuase z is to the power of 6.


Determine the complex constant c such that v is a root of
z^6 − c = 0
Plot all roots in the complex plane.
v= √3 − i
Click to expand...
Yes, there will be six roots. A general way to start would be to divide out the root you know -

(z^2 - c)/(v - √3 + i)

must have a remainder of zero. You should be able to solve for c.

If c is turns out to be real, then complex roots (of a polynomial with real coefficients!) happen in conjugate pairs, so √3 + i is also a root. The corresponding factors of the polynomial are

(v - √3 + i)(v - √3 - i) = (v - √3)^2 - (i)^2 = ...

When you divide (z^6 - c) by that product, the remainder must be zero. You could try that division first to see if you get a real c .. that would be an easier division to do.

OOPS -
 
so first when you say (v - √3 + i) do you actually mean (z - √3 + i)?

I am still very confused

so the first thing you want me to do is solve :(z^2 - c)/(v - √3 + i) to get the 'c' which is complex.

well what would the next step be? (z^2 - c)/( √3 - i- √3 + i)?


 
kmalik001 said:
so first when you say (v - √3 + i) do you actually mean (z - √3 + i)?
I am still very confused
Click to expand...


Actually I have a different take on this.
\(\displaystyle c=-64\) because \(\displaystyle v=2\exp\left(\dfrac{-\pi i}{6}\right)\). That is \(\displaystyle v^6-(-64)=0\)

Now we have one root of \(\displaystyle z^6+64=0\). let \(\displaystyle \rho = \exp \left( {\dfrac{{i\pi }}{3}} \right)\).

Consider the six numbers \(\displaystyle v\cdot\rho^k,~k=0,~1,\cdots~5\). Each of those is a root of \(\displaystyle z^6+64=0\).

Note that if \(\displaystyle k=0\) we get \(\displaystyle v=\sqrt{3}-i\).
 
2 things i don't quite get.
First where did you get 64. You merely converted v into polar form so then where did the 64 come from.
and second where did you get pi/3 from?
 
and in the question it says that c is a complex constant

Determine the complex constant c such that v is a root of
z6 − c = 0
 
kmalik001 said:
2 things i don't quite get.
First where did you get 64. You merely converted v into polar form so then where did the 64 come from.
and second where did you get pi/3 from?
Click to expand...

kmalik001 said:
and in the question it says that c is a complex constant Determine the complex constant c such that v is a root of \(\displaystyle z^6 − c = 0\)
Click to expand...



First do you understand that \(\displaystyle v^6=(\sqrt{3}-i)^6=-64~? \)
If you do then clearly \(\displaystyle v=(\sqrt{3}-i) \) is a root of \(\displaystyle z^6+64=0 \) or \(\displaystyle c=-64 \).

From you original work, c must be a real number other wise you cannot assume that if \(\displaystyle v \) is a root then \(\displaystyle \overline{v} \) is also a root,
That only works for polynomials with real coefficients.

There are six sixth roots of \(\displaystyle -64 \).
Those six numbers are all on a circle centered at the origin with radius equal \(\displaystyle 2\).
They are all equally spaced at angles of \(\displaystyle \left( \dfrac{2\pi}{6} \right) = \left(\dfrac{\pi}{3}\right) \) between them.
 
I didn't know that the conjugate also being a root was only for real number coefficients.
I understand the method that you have explained and I believe that this is what the question is asking.
But if c were to be a complex number then how would I solve it?

Thank you so much for explaining :)
 
kmalik001 said:
I didn't know that the conjugate also being a root was only for real number coefficients.
I understand the method that you have explained and I believe that this is what the question is asking.
But if c were to be a complex number then how would I solve it?

Thank you so much for explaining :)
Click to expand...

c is a complex number with imaginary part = 0
 
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