Complex roots - properHoc

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(This page leads to an interactive page.)

Complex roots

When taking z to the power of n we take the modulus to the power of n, and multiply the argument by n, So how about the n^th root? We do the opposite. We take the n^th root of the modulus and then we divide the argument by n. So, if

$\displaystyle w=c+di=q\left( {\cos \beta +i\sin \beta } \right)$

then

$\displaystyle \sqrt[n]{w}=\sqrt[n]{q}\left( {\cos \left( {\beta /n} \right)+i\sin \left( {\beta /n} \right)} \right)$

But how about solving the equation zⁿ=w in general? We have

$\displaystyle {{z}^{n}}={{p}^{n}}\left( {\cos \left( {n\alpha } \right)+i\sin \left( {n\alpha } \right)} \right)=w=c+di$

But, for w, we may add any multiple of 2π to the angle without changing the equation.

$\displaystyle w=c+di=q\left( {\cos \beta +i\sin \beta } \right)=q\left( {\cos \left( {\beta +2\pi m} \right)+i\sin \left( {\beta +2\pi m} \right)} \right)$

Next we do as before,

$\displaystyle \sqrt[n]{w}=\sqrt[n]{q}\left( {\cos \left( {\frac{{\beta +2\pi m}}{n}} \right)+i\sin \left( {\frac{{\beta +2\pi m}}{n}} \right)} \right)$

This means that we get one solution as before, thenwe may add multiples of 2π/n untill we get to 2π, then we start all over again. That means that we will get n evenly spaced solutions.

The solution with the largest real part is called the prinipal root. If two values have equal biggest real part, then the one with a positive real part is selected.

On the interactive page you may select the power, then you may move the green point representing w. The principal root is indicated by a slightly larger dot.

Link to the interactive page:

Complex roots

Up a level : Complex numbers
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Next page : Complex Quadratic equations

Last modified: Nov 10, 2025 @ 10:12