If I have a complex number of the form $z = a+bi$, how would I find the complex roots? I know that each root will be equidistant from each other and will form a circle, but I'm not sure how to solve for the complex roots. I also know that for a complex number with $n$ roots, each root will be rotated $2\pi/n$ radians. It's not an issue of graphing the complex roots, but it is for solving for them.
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Complex numbers do not have roots. Equations have roots. For example, the equation $z^4 = 1$ has roots $z = \pm 1, \pm \mathrm{i}$. Perhaps you are confusing square roots, cube roots, etc, with roots. – Fly by Night Jul 11 '15 at 22:30
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Express the number in polar form $$z=re^{i\theta} $$ and solve the equation $w^n=z$ , so that $ w^n =r'^ne^{in(\omega+2k\pi)}$ and use the fact that, in this polar form, two numbers are equal if they have the same radius and same angle.
Gary.
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You have to write $z$ in exponential form: $$a=r\mathrm e{\mathrm i \theta}$$ The $n$-th roots of $z$ are then $$ r^{\tfrac1n}\mkern1mu\mathrm e^{\mathrm i\tfrac{\theta+2k\pi}n}\quad (k=0,1,\dots n-1). $$ Note that once you one $n$-th root, all other roots are obtained multiplying it by the $n$-th roots of $1$.
Bernard
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