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murshid_islam
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what does [tex]\sqrt{i^2}[/tex] equal to? is [tex]\sqrt{i^2} = i[/tex] or [tex]\sqrt{i^2} = \pm i[/tex]?
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you said, [itex]\sqrt{z} = z^{1/2} = \sqrt{r}e^{i\theta/2}.[/itex]Data said:I gave you the usual definition of the square root function in your other thread. You can use that to figure out what it is.
but if [tex]z = x + iy[/tex], then [tex]r = \sqrt{x^2 + y^2}[/tex] and [tex]\theta = \tan^{-1}\left( y \over x \right)[/tex]Data said:So let's try out this definition. If [itex]z=-1[/itex], then we write [itex]z = e^{i\pi}[/itex], and we get [itex]\sqrt{z} = e^{i\pi / 2} = i[/itex]. As you might expect.
murshid_islam said:what does [tex]\sqrt{i^2}[/tex] equal to? is [tex]\sqrt{i^2} = i[/tex] or [tex]\sqrt{i^2} = \pm i[/tex]?
murshid_islam said:you said, [itex]\sqrt{z} = z^{1/2} = \sqrt{r}e^{i\theta/2}.[/itex]but if [tex]z = x + iy[/tex], then [tex]r = \sqrt{x^2 + y^2}[/tex] and [tex]\theta = \tan^{-1}\left( y \over x \right)[/tex]
The square root of i^2 is -i. This can be simplified to just -i, as i^2 is equivalent to -1.
To find the square root of i^2, you can use the property that the square root of a complex number can be found by taking the square root of its magnitude and dividing the argument by 2. In this case, the magnitude of i^2 is 1, and the argument is 180 degrees (or pi radians). So the square root of i^2 is sqrt(1) * e^(i * 180 / 2), which simplifies to -i.
Yes, the square root of i^2 will always be -i. This is because i^2 is equivalent to -1, and the square root of -1 is always a complex number with a magnitude of 1 and an argument of 180 degrees (or pi radians). Therefore, the square root of i^2 will always be -i.
No, the square root of i^2 cannot be simplified any further. It is already in its simplest form, which is -i.
The square root of i^2 is equal to -i because i^2 is equivalent to -1. The square root of -1 is always a complex number with a magnitude of 1 and an argument of 180 degrees (or pi radians). This complex number can be expressed as -i, so the square root of i^2 is -i.