Find the square roots of i

?

would the modulus = i ? and not sure how to work out its argument?

any help apprecited

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- Nov 21st 2013, 03:23 AMTweetysquare root of i
Find the square roots of i

?

would the modulus = i ? and not sure how to work out its argument?

any help apprecited - Nov 21st 2013, 03:36 AMPlatoRe: square root of i
- Nov 21st 2013, 03:37 AMSlipEternalRe: square root of i
$\displaystyle e^{i\theta} = \cos \theta + i \sin \theta$

Plug in $\displaystyle \dfrac{\pi}{2} + 2\pi n$:

$\displaystyle e^{i \left(\pi/2 + 2\pi n\right)} = i$

Take the square root of both sides:

$\displaystyle e^{i\left(\pi/4 + \pi n\right)} = \sqrt{i}$

There are two solutions in the range $\displaystyle [0,2\pi)$:

$\displaystyle e^{\pi i/4},e^{5\pi i/4}$ - Nov 21st 2013, 03:50 AMTweetyRe: square root of i
why plug in pi/2 ?

- Nov 21st 2013, 04:00 AMTweetyRe: square root of i
why cant I just work it out like this, the above looks quite complicated,;.....

let z = 0 + 1*i

than the modulus of z = $\displaystyle \sqrt{0^{2} + 1^{2}} = 1 $

arg(z) = $\displaystyle tan \alpa = 1 = \frac{\pi}{4}} $

but not sure how to go from here, the correct answers are

$\displaystyle \frac{1}{\sqrt{2}}(1+i) $

$\displaystyle -\frac{1}{\sqrt{2}}(1+i) $ - Nov 21st 2013, 07:02 AMHallsofIvyRe: square root of i
First, what made you think, originally, that there were

**four**square roots of i?

Second, while the modulus of z= i is 1, its argument is NOT [itex]\pi/4[/itex]. I don't know where you got that "1" in "$\displaystyle arctan(1)$". Surely you don't think argument is the arctangent of the**modulus**? The argument of "a+ bi" is the arctangent of b/a, if a is not 0. Of course, if a= 0 then a+ bi= bi which has modulus $\displaystyle \frac{\pi}{2}$ if b> 0, $\displaystyle -\frac{\pi}{2}$ if b< 0. (0 has no "argument").

So if z= i, $\displaystyle arg(z)= \frac{\pi}{2}$. HALF of that is $\displaystyle \frac{\pi}{4}$ and half of $\displaystyle \frac{\pi}{2}+ 2\pi= \frac{5\pi}{2}$ is $\displaystyle \frac{5\pi}{4}$. Those are the arguments of the two square roots of i which, of course, have modulus $\displaystyle 1^{1/2}= 1$, so the two square roots are $\displaystyle \frac{\sqrt{2}}{2}(1+ i)$ and $\displaystyle \frac{\sqrt{2}}{2}(1- i)$.