Roots of an equation

**arbolis** · August 12th 2009, 10:45 AM

I must find the root of $z^3+1=0$ .
I both used the polar and Cartesian approaches and reached the result that the root is $z=-1$ .
Is it possible? Shouldn't there be 3 roots? (they could be multiple roots I guess).
I can show the details of my attempt if you like.

**running-gag** · August 12th 2009, 10:50 AM

You have found only one root !
Now you can factor $z^3+1 = (z+1)(z^2-z+1)$ and solve $z^2-z+1=0$ to find the two other roots

**running-gag** · August 12th 2009, 10:56 AM

Anyway you should have found 3 roots

$z^3 = -1$

$r^3 \:e^{3i \theta} = e^{i \pi}$

$r^3=1$ and $3 \theta = \pi + 2k \pi$

$r=1$ and $\theta = \frac{\pi}{3} + \frac{2k \pi}{3}$

$z_1 = e^{i \frac{\pi}{3}}$

$z_2 = e^{i \pi} = -1$

$z_3 = e^{i \frac{5 \pi}{3}} = e^{-i \frac{\pi}{3}}$

Thread: Roots of an equation

LinkBack

Thread Tools

Display

Roots of an equation

Similar Math Help Forum Discussions

Can this equation have any other roots

Roots on an equation

roots of the equation

roots of equation

Roots of equation

Search Tags