z8 + 1 = 0

[itsangelan]

Hi guys, just wondering if anyone can help me with this answering this question.

Find the solutions of the equation z8 + 1 = 0 in polar form. Hence factorise z8 + 1.

I've firstly found the first 4 solutions where z=cisπ/8, cis3π/8, cis5π/8 and 7π/8.
However the other 4 solutions are greater than π which are 9π/8, 11π/8, 13π/8 and 15π/8. I thought the domain was between π and -π so why are the solutions so?

[Syndicate]

Hi guys, just wondering if anyone can help me with this answering this question.

Find the solutions of the equation z8 + 1 = 0 in polar form. Hence factorise z8 + 1.

I've firstly found the first 4 solutions where z=cisπ/8, cis3π/8, cis5π/8 and 7π/8.
However the other 4 solutions are greater than π which are 9π/8, 11π/8, 13π/8 and 15π/8. I thought the domain was between π and -π so why are the solutions so?

You are absolutely correct saying that the other four solutions are not part of the (-Pi, Pi] convention, however, they can be.

You can convert the following arguments into principal arguments by subtracting by 2pi.

So 9pi/ 8 becomes -7pi/8, 11pi/8 becomes -5pi/8, 13pi/8 becomes -3pi/8 and 15pi/8 becomes -Pi/8.

This skill is quite necessary, and it's always better to list your arguments as principal arguments.

[Shadowxo]

Hi guys, just wondering if anyone can help me with this answering this question.

Find the solutions of the equation z8 + 1 = 0 in polar form. Hence factorise z8 + 1.

I've firstly found the first 4 solutions where z=cisπ/8, cis3π/8, cis5π/8 and 7π/8.
However the other 4 solutions are greater than π which are 9π/8, 11π/8, 13π/8 and 15π/8. I thought the domain was between π and -π so why are the solutions so?

Like Syndicate said,
the solutions can be between (-pi,pi] meaning they can be negative
so solutions are 8theta = -7pi, -5pi, -3pi, -pi, pi, 3pi, 5pi, 7pi
In order to find all the solutions usually around half will be negative

[lzxnl]

Note the geometric property of the solutions to the equation ; they're distributed evenly about a circle of size . If you are to fit n solutions in a circle of angle 2pi, then each solution must be off by 2pi/n in argument. Therefore, it suffices to find one solution and then keep adding/subtracting 2pi/n until you start repeating solutions. This will happen when you have exactly n solutions. So, start from the simplest solution, which is cis(pi/8), and add/subtract 2pi/8 = pi/4 to the argument. Generally, you'll find that it's easiest to construct 4 solutions by successively adding pi/4 and the other 4 by successively subtracting pi/4 from pi/8.