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Find the polar form of the roots of the polinomial

$p(z)=iz^2-2z+1+2i$

I found the roots $z_1=-i-\sqrt{i-3}$ and $z_2=-i+\sqrt{i-3}$ but I don't know how to deal with the complex numbers inside the square root in order to isolate both real and imaginary parts.

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    $\begingroup$ Are you sure you typed the roots correctly? $z_1=z_2$. $\endgroup$ – Parcly Taxel Mar 28 '17 at 15:44
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    $\begingroup$ I corrected, thanks $\endgroup$ – AnalyticHarmony Mar 28 '17 at 15:54
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To compute the square root of a complex number: $(x+iy)^2=-3+i$, expand the l.h.s. and identify the real and imaginary parts: $$x^2-y^2=-3, \quad xy=\frac12.$$ You simplify the computation observing the square of the modulus of $x+iy$ is the modulus of $-3+i$: $$x^2+y^2=\sqrt{10}.$$ So we have a linear system in $x^2$ and $y^2$: $$\begin{cases}x^2+y^2=\sqrt{10}\\x^2-y^2=-3\end{cases}\iff\begin{cases}x^2=\dfrac{\sqrt{10}+3}2\\ y^2=\dfrac{\sqrt{10}-3}2\end{cases}$$ Observe $xy>0$, so $x$ and $y$ have the same sign, and ultimately $$x+iy=\pm\frac12\biggl(\sqrt{2\sqrt{10}+6}+i\sqrt{2\sqrt{10}-6}\biggr).$$

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Given $y = a x^2+bx + c$, the roots are given by the quadratic formula $$ y_{\pm} = \frac{-b \pm \sqrt{b^2 - 4 a c}}{2a}. $$ Here $$ a = i, \quad b = -2, \quad c = 1 + 2i, $$ therefore $$ z_{\pm} = -i \left(1 \pm \sqrt{3 - i} \right). $$ Compute the magnitude and argument.

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  • $\begingroup$ My question, as the title states, is concerned with the transformation to polar form, given a number in the form of the roots obtained. The magnitude, as far as I know, is given by $\sqrt{Re(z)^2+Im(z)^2}$, but I don't know how to find $Re(z)$ and $Im(z)$. $\endgroup$ – AnalyticHarmony Mar 28 '17 at 16:12

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