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 Re and Im Radical Expression of square root of $a+ib, \quad (a,b \in \mathbb{R})$
12-18-2010, 03:08 PM
Post: #1
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Re and Im Radical Expression of square root of $a+ib, \quad (a,b \in \mathbb{R})$
If $x,y \in \mathbb{R}$ such that $(x+iy)^2=a+ib$ then $x^2-y^2=a$ and $2xy = b$ and so when $b != 0$
$\displaystyle{y = \frac{b}{2x}, \quad x^2 - \left(\frac{b}{2x}\right)^2 = a}$ and so $4 x^4 - 4a x^2 - b^2 = 0$ therefore
$\displaystyle{x^2 = \frac{a + \sqrt{a^2 + b^2}}{2}}$ is the only positive solution. From $\displaystyle{\frac{b}{2x}}$ we get
$\displaystyle{x + iy = \pm \frac{2}{\sqrt{2}} \left(\sqrt{a+\sqrt{a^2+b^2})}+ i \frac{b}{|b|} \sqrt{-a+\sqrt{a^2+b^2}}\right)}$
and the general formula follows:
$\displaystyle{x + iy = \pm \frac{1}{\sqrt{2}} \left(\sqrt{a+\sqrt{a^2+b^2}}+ i\; \text{sgn} (b+) \sqrt{-a+\sqrt{a^2+b^2}}\right)}$
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