# Thread: square root of i

1. ## square 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

2. ## Re: square root of i

Originally Posted by Tweety
Find the square roots of i ?
Well of course as with any complex number there are two square roots:
$\sqrt {|z|} \exp \left( {\frac{{\arg (z)i}}{2} + \pi ki} \right),~k=0,~1$

3. ## Re: square root of i

$e^{i\theta} = \cos \theta + i \sin \theta$

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

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

Take the square root of both sides:

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

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

$e^{\pi i/4},e^{5\pi i/4}$

4. ## Re: square root of i

why plug in pi/2 ?

5. ## Re: 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 = $\sqrt{0^{2} + 1^{2}} = 1$

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

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

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

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

6. ## Re: 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 $\pi/4$. I don't know where you got that "1" in " $arctan(1)$". 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 $\frac{\pi}{2}$ if b> 0, $-\frac{\pi}{2}$ if b< 0. (0 has no "argument").

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