Thread: Find six distinct sixth roots of -1 + i roo3

1. Find six distinct sixth roots of -1 + i roo3

Hi

I want ckeck my answer :

Q 1: see picture ( I want check )

Q2 : Show that sum of cube root of unyy is zero ( i want explaine the soving of it )

2. The way you have written this, it is very hard to tell what you have done.

You have correctly changed the the number to polar form: $\displaystyle -1+ i\sqrt{3}= 2e^{2i\pi/3}$.

To find the first 6th root, you take the 6th root of 2 and divide $\displaystyle 2\pi/3$ by 6: $\displaystyle 2^{1/6}e^{i\pi/9}$. To find the second root add $\displaystyle 2i\pi$ to the original polar form: $\displaystyle 2e^{2i\pi/3+ 2i\pi}= 2e^{8i\pi/3}$. since $\displaystyle e^x$ has period $\displaystyle 2i\pi$ that is the same number but dividing the argument by 6 gives $\displaystyle 2^{1/6}e^{4i\pi/9}$ gives a new root.

Keep $\displaystyle 2i\pi$ and then divide by 6 to get the other 4 roots.

Geometrically, since the distance from the origin is always $\displaystyle 2^{1/6}$ and the angle is always additional $\displaystyle 2\pi/6= 2\pi/3$, these numbers are the vertices of a regular hexgon in the complex plane. Since one vertex is alway exactly opposite another vertex, it should be easy to see what the sum of the roots is even without doing any arithmetic. An algebraic way of seeing it is to remember that when you multiply things like $\displaystyle (x- a)(x- b)...(x- z)$, the coefficient of x is just a+ b+ c+ ...+ z/

3. I accept with information:
You have correctly changed the the number to polar form: .

To find the first 6th root, you take the 6th root of 2 and divide by 6: . To find the second root add to the original polar form: . since has period that is the same number but dividing the argument by 6 gives gives a new root.

Keep and then divide by 6 to get the other 4 roots.

__________________________
Watch The Losers Online Free
Watch The Back-Up Plan Online Free

4. I try to solve agine

I try to get K1 and k2 is correct my answer >>

5. In K1, you want the exponent to be $\displaystyle \frac{2i\pi}{3}\frac{1}{6}$, not "$\displaystyle \frac{2i\pi}{6}$" as you have. However, then you have $\displaystyle \frac{i\pi}{9}$ which is correct. Also, you should NOT write "360" when you mean $\displaystyle 2\pi$.

For K2, you have exponent $\displaystyle \frac{12\pi}{27}$ which is [b]almost[/tex] correct- you forgot the "i". Also it can be reduced: $\displaystyle \frac{12\pi i}{27}= \frac{4\pi i}{9}$. (Surely you know to get the least common divisor when adding fractions?)