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I have the following quadratic equation:

$$(1-i)x^2 +ix - \frac{1}{4}i = 0$$

when I tried to solve for $x$ and get the complex solutions I got $ \sqrt{i}$ in the determinant.

I know that

$$ i^2 = -1 \implies \sqrt{i} = (-1) ^{\frac{1}{4}}$$ but don't come any further with it.

Thanks in advance for your help

Ozk
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    A good place to start when taking square roots of complex numbers is to convert them into polar form, e.g. $i = e^{i \pi/2}$. Of course, $i$ has multiple square roots, like any other number. – jwimberley Jan 26 '21 at 14:39

1 Answers1

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Note that $i=e^{\pi i/2}$. Therefore, the square roots of $i$ are $\pm e^{\pi i/4}$; in other words, they are$$\pm\left(\frac1{\sqrt2}+\frac i{\sqrt2}\right).$$You can reach the same conclusion solving the system$$\left\{\begin{array}{l}a^2-b^2=0\\2ab=1\end{array}\right.$$and taking $a+bi$.

José Carlos Santos
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