When I type in the sqrt of 'i' in the calculator.....?

Question

When I type rt(i) in the calculator, it gives me a weird answer that is not related to i or -1 (it gives me the answer ".7071067812+.707...."). How is this so? How can you take the sqrt of the sqrt of -1 (i) and get this sort of answer?

Answer 1

No.

Answer 2

The square root of a number a is by definition a number z such that

z^2 = a

In your case, the calculator tried to find a number z whose square is i:

z^2 = i

There are two such numbers, and they are opposite each other (z1 = -z2). The number generated by your calculator is precisely

z = (1 + i) / (sqrt 2)

Check it out:

z^2 = (1 + i)^2 / (sqrt 2)^2
= (1+i)(1+i) / 2
= (1 + i + i + i^2) / 2
= (1 + 2i - 1) / 2 = i

Answer 3

If you scroll to the right (push the arrow right button) after you ask for the answer to sqrt (i), I belive you will see an i after the numbers. You're calcualtor is showing what the value would be in imaginary notation. You can actually turn this off on many models if you see fit, but it will just say that there is an error in the domain or some such message.

Answer 4

you're asking for the fourth root of -1. Either your calculator can't handle that math, or the answer given is also an imaginary number, or a complex number.

Answer 5

the roots of imaginary numbers can be shown in this way

recall:
e^(i*pi)=-1 so
e^(i*t)=cos(t) + i*sin(t) and is known as the Euler formula.
It is quite useful in solving 2nd order differential equations

Answer 6

cant take the square of an imaginary number

Answer 7

u cant
sq. root of -1 does not exist

Answer 8

i = cos(pi/2) + i sin(pi/2) = e^(i*pi/2)

sqrt(i) = sqrt{e^(i*pi/2)} = e^(i*pi/4)

= cos(pi/4) + i sin(pi/4)

= 1/sqrt2 + i/sqrt2

= 0.7071067812 + i (0.7071067812)

Answer 9

exactly the way qwert has done it.

Answer 10

qwart has explained it correctly.