The question:

Find the roots (in Cartesian Form) of

I'm not sure where to start. I tried applying the quadratic formula, but it doesn't seem to help. Any help would be great.

Printable View

- Aug 14th 2010, 04:15 AMGlitchFinding the roots of a complex number
The question:

Find the roots (in Cartesian Form) of

I'm not sure where to start. I tried applying the quadratic formula, but it doesn't seem to help. Any help would be great. - Aug 14th 2010, 04:39 AMVlasev
By the quadratic formula you should get

[LaTeX ERROR: Compile failed]

From here you need to use that the square root is multivalued. - Aug 14th 2010, 05:10 AMHallsofIvy
It is simplest to take roots by using DeMoivre's formula: The nth roots of complex number , in polar form, are given by where k= 0 to n-1 give the n roots. (r must be positive of course, and is the positive real n th root.)

In particular, if [itex]a+ bi= r e^{i\theta}= r(cos(\theta)+ i sin(\theta))[/itex] then the square root of a+ bi is given by and - Aug 14th 2010, 05:28 AMmr fantastic
An alternative approach to the excellent approaches already suggested is to substitute z = a + ib, expand and equate the real and imaginary parts to zero. This gives two simultaneous equations to be solved for a and b, leading to the two solutions z = 2 + i and z = 1 - i.

- Aug 14th 2010, 07:02 AMGlitch
- Aug 14th 2010, 02:49 PMVlasev
Yea. You can do that. You will get two solutions (1+2i) and (-1-2i). Plugging both in the formula above, you will get a total of 2 different solutions.