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When finding n-th root of the complex number z=cos(x)+sin(x)i, we can use known De Moivre's formula:

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My task is to find enter image description here without using De Moivre's formula, and my question is how to do that?

Thank you for your answers.

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    $\begingroup$ you could use Euler's form instead, $z=|z|e^{arg(z)\cdot i}$ in which case $z^{\frac{1}{n}} = |z|^{\frac{1}{n}}e^{arg(z)\cdot i / n}$, but in reality that is just demoivre's formula in disguise. $\endgroup$ – JMoravitz Nov 27 '14 at 20:32
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Note that if the root is $a+bi$ then $a^2-b^2=3$ and $2ab=2$

Then $a^2b^2=1$ so that $(a^2-b^2)^2+4a^2b^2=(a^2+b^2)^2=13$ and $a^2+b^2=\sqrt {13}$

Then it is easy to solve for $a^2$ and $b^2$, bearing in mind that $a$ and $b$ are real. Then you get $a$ and $b$, and fix the signs of the square roots to make the product come out with the right sign.

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Here's a technique that I can only recommend because you asked this question. Suppose that $$\sqrt{3+2i} = a + b\,i$$ or, equivalently, $$3+2\,i = (a+b\,i)^2 = (a^2-b^2) + (2ab)i.$$ Equating real and imaginary parts, we get \begin{align} a^2-b^2 &= 3 \\ 2ab &= 2. \end{align} Thus, you can find $a$ and $b$ by solving this system. From the first equation, we have $a^2=b^2+3$; you can plug that back into the second equation to get a quartic that can, in principle, be solved using Tartaglia's method. Applying Mathematica to this system, I found that $$a+bi= \pm \left(\sqrt{\frac{1}{2} \left(3+\sqrt{13}\right)}+\frac{1}{2} i \left(\sqrt{13}-3\right) \sqrt{\frac{1}{2} \left(3+\sqrt{13}\right)}\right).$$

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    $\begingroup$ This is really elegant :) Thank you! $\endgroup$ – Mr.M Nov 27 '14 at 21:09

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