Square root of complex number -2i

So I was given that z=(6-8i)/(4+3i), and I need to find the square root of z. I simplify z to -2i then I am not sure how to proceed. Do I need to convert the expression into exponential form first? Thank you so much!

You don't need to: sqrt(-2i) = sqrt(-1) * sqrt(2) * sqrt(i)
The square root of -1 is i, up to a sign
The square root of 2 is already real
The square root of i
- has a unit length
- has the argument 45 degrees, or pi/4
- this makes it equal to (1+i)/sqrt(2)
Multiplying the three terms we get
i * sqrt(2) * (1+i)/sqrt(2)
= i * (1+i)
= i-1

Square root of i = 1/sqrt(2) + i / sqrt(2)
Square root of -2 = i sqrt(2)

square root of -2i = i sqrt(2) * (1 / sqrt(2) + i / sqrt(2) ) = -1 + i

Since this is the negative square root (the real part is negative), multiply by -1 to get the positive square root: 1 - i.

sqr(-2i) = sqrt(-2)sqrt(i) = i*sqrt(2)*e^(ipi/4)

Recall that cos(pi/2) + isin(pi/2) = i and that (cos(pi/4) + isin(pi/4)) = sqrt(i)

√ (-2i) = √ -2 * √ i = i√ 2 * e^(i*pi/4)