# Complex Solutions

1. Jan 10, 2009

### salman213

1. Hi I posted a previous question on this forum and it was answered well but I had another question about complex solutions/roots.

for example if i have a question like

z^2 = 1e^(j)(pie)

z = 1e^(j)(pie + 2kpie)^1/2 k =0,1

1. z = 1e^(j)(pie/2 ) = 0 + j
2. z = 1e^(j)(3pie/2) = 0 - j

if I test these solutions (0+j)(0+j) = -1 , (0 - j)(0 - j) = -1

they are correct but my question is are those angles correct????

another way to solve that question seems to be

z^2 = 1e^(j)(-pie)

z = 1e^(j)(-pie + 2kpie)^1/2 k =0,1

1. z = 1e^(j)(-pie/2 ) = 0 - j
2. z = 1e^(j)(pie/2) = 0 + j

if i go backwards

0 + j = 1e^(j)(-pie/2 ) and 0 - j = 1e^(j)(pie/2)

so which angles are correct???

Like on an exam I don`t really know which angles I would write!

2. Jan 10, 2009

### gabbagabbahey

Both angles are correct; the complex exponential is periodic with period 2pi, so if you add 2pi to your angle of -pi/2, you get the same value for the exponential

i.e. $$e^{j\frac{-\pi}{2}}=e^{j\left(\frac{-\pi}{2}+2\pi\right)}=e^{j\frac{3\pi}{2}}=-j$$