What is the square root of -i?

Answer 1

just like sqrt(-1) is not defined in the real numbers, sqrt(-i) isn't defined in the complex numbers. for this to exist, you need to be in a field where it is defined like that. a system of numbers where a number X = a + bi + cj and j x j = -i
the thing with algebra is that you can simply state that fact and BAM, you have another algebraic space, and you can just say that sqrt(-i) = j

Answer 2

Solve (a+bi)² = -i
→ a² + 2abi - b² = -i

This yields the equations for a and b...
a² - b² = 0 (real part)
2ab = -1 (imaginary part) → b = -1/2a
→ a² - (1/2a)² = 0
→ a² = 1/4a²
→ a^4 = 1/4
→ a = 1/sqrt2, b = -1/sqrt2

So sqrt(-i) = ± (1 - i) / sqrt2.

Answer 3

You have to use Euler's equation.

e^(ix) = cos (x) + i sin (x)

Where -i would be x = (3/2)pi

e^(i(3/2)pi) = -i

sqrt of a number is the same thing as raising it to the 1/2 power.

So squrt of -i would be e^(i (3/4) pi)

e^(i(3/4)pi) = cos(pi 3/4) + isin(pi 3/4)
= -0.707 + i(0.707)
= 0.707(i-1)

Answer 4

I don't think that exists.

Anything square root of a negative number doesn't exist...

Answer 5

i as in the imaginary number?

well i = sqrt(-1)

for square root of that is the sqrt(sqrt(-1))
or the fourth root of -1

Answer 6

Okay, -i is defined as being the square root of -1. So, it goes that (-i) is minus(square root of minus one). just put a minus sign in front of the square rootbracket of -1.

Answer 7

√(-i) = ±(i-1)√2/2

Answer 8

there isn't one, is there? b/c i stands for imaginary.