Are all solutions of x^4=-1 in the complex plane valid?

[ques1988]

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?

 

[phyzguy]

Yes, it has a solution. In fact, the fundamental theorem of algebra tells you that it has 4 solutions. In this case they are :
$$\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i, -\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i,\frac{\sqrt2}{2}-\frac{\sqrt2}{2}i,-\frac{\sqrt2}{2}-\frac{\sqrt2}{2}i$$

 

[abhishek ghos]

yup,it is valid
x= +/- sqrt(i)
sqrt(i) = +/- (1+i)/sqrt(2)
so you get four solutions

 

[hkBattousai]

Yes, it has a solution. In fact, the fundamental theorem of algebra tells you that it has 4 solutions. In this case they are :
$$\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i, -\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i,\frac{\sqrt2}{2}-\frac{\sqrt2}{2}i,-\frac{\sqrt2}{2}-\frac{\sqrt2}{2}i$$

We can also write them in exponential form.

$-1 = e^{j180} = e^{j540} = e^{j900} = e^{j1260}$

Solutions are:
$e^{j\frac{180}{4}} = e^{j45} = +\frac{\sqrt2}{2}+j\frac{\sqrt2}{2}$
$e^{j\frac{540}{4}} = e^{j135} = -\frac{\sqrt2}{2}+j\frac{\sqrt2}{2}$
$e^{j\frac{900}{4}} = e^{j225} = e^{-j135} = -\frac{\sqrt2}{2}-j\frac{\sqrt2}{2}$
$e^{j\frac{1260}{4}} = e^{j315} = e^{-j45} = +\frac{\sqrt2}{2}-j\frac{\sqrt2}{2}$

 

[blue_raver22]

I would like to clarify that the equation x^4=-1 does not have a real solution. However, in the complex plane, it does have four complex solutions. One of these solutions is indeed sqrt(i), which can be written as ±(1+i)/√2. This solution is valid and can be verified by substituting it back into the original equation. However, it is important to note that the other three solutions are also valid and should not be ignored. These solutions are -sqrt(i), ±(1-i)/√2, and -sqrt(i). In the complex plane, all four solutions satisfy the given equation and are equally valid. Therefore, it is important to consider all solutions when solving equations in the complex plane.