how do you put the following complex numbers in polar and engineering format?

Question

z = root of j
where j = (-1)

sorry i meant square root of -1

Answer 1

Hi. I'm not so sure what engineering format is, but polar format can be found by using

r exp(iφ) = r (cos(φ) + isin(φ))

now say z = sqrt(-1) = i, so φ=90 deg = π/2 and r=1
or i = exp(i π/2)

I guess you mean all roots,I'm not quite sure...

(-1)^(1/n) = (-1)^(1/2 x 2/n) = i^(2/n) = exp(i π/n)

Answer 2

The square root of minus 1 is i. In polar format, that's r=1, theta = pi/2.