- #1

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## Main Question or Discussion Point

I am trying to find a sollution to this in the complex plane. One that seems to work is sqrt(i), but is this valid or not?

- Thread starter ques1988
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- #1

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I am trying to find a sollution to this in the complex plane. One that seems to work is sqrt(i), but is this valid or not?

- #2

phyzguy

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[tex]\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i, -\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i,\frac{\sqrt2}{2}-\frac{\sqrt2}{2}i,-\frac{\sqrt2}{2}-\frac{\sqrt2}{2}i[/tex]

- #3

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yup,it is valid

x= +/- sqrt(i)

sqrt(i) = +/- (1+i)/sqrt(2)

so you get four solutions

x= +/- sqrt(i)

sqrt(i) = +/- (1+i)/sqrt(2)

so you get four solutions

- #4

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We can also write them in exponential form.

[tex]\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i, -\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i,\frac{\sqrt2}{2}-\frac{\sqrt2}{2}i,-\frac{\sqrt2}{2}-\frac{\sqrt2}{2}i[/tex]

[itex]-1 = e^{j180} = e^{j540} = e^{j900} = e^{j1260}[/itex]

Solutions are:

[itex]e^{j\frac{180}{4}} = e^{j45} = +\frac{\sqrt2}{2}+j\frac{\sqrt2}{2}[/itex]

[itex]e^{j\frac{540}{4}} = e^{j135} = -\frac{\sqrt2}{2}+j\frac{\sqrt2}{2}[/itex]

[itex]e^{j\frac{900}{4}} = e^{j225} = e^{-j135} = -\frac{\sqrt2}{2}-j\frac{\sqrt2}{2}[/itex]

[itex]e^{j\frac{1260}{4}} = e^{j315} = e^{-j45} = +\frac{\sqrt2}{2}-j\frac{\sqrt2}{2}[/itex]

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