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I have expressed a complex polynomial in terms of its six linear factors, as $(z - x_0)(z - x_1)$ etc., where each x is a root in exponential polar form.

I have been able to express one conjugate pair of factors instead using cartesian form for the root. I can then multiply out that pair of factors and express the result in a simpler form.

The other pairs however I do not know how to express in cartesian form, as I am unable to determine the real and imaginary values from the exponential form.

Is there a formula/shortcut/simpler way of multiplying out linear factors corresponding to conjugate pairs of roots while the roots are in exponential form?

Otherwise my result is a large and seemingly messy number of terms that must continue to be multiplied out.

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  • $\begingroup$ An example would help. It's not clear what you mean by express one conjugate pair of factors. $\endgroup$ – dxiv Mar 26 '17 at 6:19
  • $\begingroup$ Apologies @dxiv. I meant that out of the three pairs of factors (which corresponded to conjugate pairs of roots) I could express only one of them easily in cartesian form. This meant I could use $(z - a)(z - \overline a) = z^2 - 2xz + (x^2 + y^2)$. However, I was unable to simplify the remaining pairs despite knowing that they comprised of conjugates. Sorry, I am new to math notation so was not confident in providing the details well via notation. $\endgroup$ – remnant Mar 26 '17 at 11:37
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For conjugate pairs, \begin{align*} (z-re^{i\theta})(z-re^{-i\theta}) &= z^2-zre^{i\theta}-zre^{-i\theta} + r^2\\ &= z^2-zr(e^{i\theta}+e^{-i\theta}) + r^2\\ &= z^2-zr(\cos\theta+i\sin\theta+\cos\theta-i\sin\theta) + r^2\\ &= z^2-2zr\cos\theta + r^2\\ \end{align*} Which is purely real, as expected. We can take this further if $re^{i\theta} = a+ib$: \begin{align*} z^2-2zr\cos\theta + r^2 &= z^2-2zr\frac{a}{r} + a^2 + b^2\ \mathrm{by\ converting\ to\ cartesian\ form}\\ &= z^2 - 2za + a^2 + b^2\\ &= (z - a)^2 + b^2\ \mathrm{by\ completing\ the\ square} \end{align*}

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    $\begingroup$ Thank you, the post-square-completion looks really nice now! $\endgroup$ – Harambe Mar 26 '17 at 7:15
  • $\begingroup$ Thank you @Shanye2020,I did not see the equivalence to polar form. This is the simplification I required. $\endgroup$ – remnant Mar 26 '17 at 11:30

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