Does x^4=-1 have a sollution

  • Thread starter ques1988
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  • #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?
 

Answers and Replies

  • #2
phyzguy
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Yes, it has a solution. In fact, the fundamental theorem of algebra tells you that it has 4 solutions. In this case they are :
[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
yup,it is valid
x= +/- sqrt(i)
sqrt(i) = +/- (1+i)/sqrt(2)
so you get four solutions
 
  • #4
Yes, it has a solution. In fact, the fundamental theorem of algebra tells you that it has 4 solutions. In this case they are :
[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]
We can also write them in exponential form.

[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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