Find six distinct sixth roots of -1 + i roo3

**r-soy** · April 21st 2010, 11:47 PM

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 )

**HallsofIvy** · April 22nd 2010, 02:43 AM

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: $-1+ i\sqrt{3}= 2e^{2i\pi/3}$ .

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

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

Geometrically, since the distance from the origin is always $2^{1/6}$ and the angle is always additional $2\pi/6= 2\pi/3$ , 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 $(x- a)(x- b)...(x- z)$ , the coefficient of x is just a+ b+ c+ ...+ z/

**jeanlee411** · April 22nd 2010, 10:08 PM

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.

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**r-soy** · April 23rd 2010, 08:40 AM

I try to solve agine

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

**HallsofIvy** · April 23rd 2010, 11:19 AM

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

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

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