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, 04:23 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, 04:36 AMPlatoRe: square root of i
- Nov 21st 2013, 04:37 AMSlipEternalRe: square root of i

Plug in :

Take the square root of both sides:

There are two solutions in the range :

- Nov 21st 2013, 04:50 AMTweetyRe: square root of i
why plug in pi/2 ?

- Nov 21st 2013, 05:00 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 =

arg(z) =

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

- Nov 21st 2013, 08:02 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 " ". 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 if b> 0, if b< 0. (0 has no "argument").

So if z= i, . HALF of that is and half of is . Those are the arguments of the two square roots of i which, of course, have modulus , so the two square roots are and .