Complex number

**matsci0000** · August 13th 2009, 09:16 PM

Find the square roots of the following complex numbers:
a)- i
b) x+ i(x^4+x^2+1)^1/2

Thanks

**pickslides** · August 13th 2009, 09:25 PM

Originally Posted by matsci0000

Find the square roots of the following complex numbers:
a) i

well $i = \sqrt{-1}$ so therefore

$\sqrt{i} = \sqrt{\sqrt{-1}} = \sqrt[4]{-1}$

**red_dog** · August 13th 2009, 10:44 PM

a) Let $z=-i$ . We have to find the complex numbers $w=a+bi$ such as $z=w^2$

$-i=a^2-b^2+2abi$ and $|z|=|w|^2$ . Then

$\left\{\begin{array}{ll}a^2-b^2=0\\2ab=-1\\a^2+b^2=1\end{array}\right.$

From the first and the tird equation we get (using the sign for a and b from the second equation)

$a=\pm\frac{\sqrt{2}}{2}, \ b=\mp\frac{\sqrt{2}}{2}$

Then, $w_1=\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i, \ w_2=-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$

b) Using the same method we have

$\left\{\begin{array}{ll}a^2-b^2=x\\2ab=\sqrt{x^4+x^2+1}\\a^2+b^2=x^2+1\end{arr ay}\right.$

We get $a=\pm\sqrt{\frac{x^2+x+1}{2}}, \ b=\pm\sqrt{\frac{x^2-x+1}{2}}$

Then, $w_1=\sqrt{\frac{x^2+x+1}{2}}+i\sqrt{\frac{x^2-x+1}{2}}, \ w_2=-\sqrt{\frac{x^2+x+1}{2}}-i\sqrt{\frac{x^2-x+1}{2}}$

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