Question

Complex numbers: z1^(1/z2) ?

Answer 1

\[(-1151570.113 - j911120.2211)^{\frac{ 1 }{ \frac{ 5\sqrt{3} }{ 2 } - j2.5 }}\]

Answer 2

1/z will be another complex number say 'y'

so its same as z^y
which we discussed earlier

Answer 3

if \[(z1)^{\frac{ 1 }{ z2 }} = e ^{\frac{ xlnr+\theta(y) }{ x ^{2}+y ^{2} }}e ^{j(\frac{ -ylnr+\theta(x) }{ x ^{2}+y ^{2} })}\]

Answer 4

\[=e ^{\frac{ \frac{ 5\sqrt{3} }{ 2 }\ln1468418.803+\theta(-2.5) }{ (\frac{ 5\sqrt{3} }{ 2 })^{2}+(2.5)^{2} }}e ^{j(\frac{ -(-2.5)\ln1468418.803+\theta(\frac{ 5\sqrt{3} }{ 2 }) }{ (\frac{ 5\sqrt{3} }{ 2 })^{2}+(2.5)^{2} })}\]

what value of theta should i use? how to find for the value of theta?

Answer 5

i am guessing
\(z_2 = x+iy \\ z_1 = r \angle \theta \)

so
\(\Large \theta = arctan(\dfrac{-911120}{-1151570})\)

Answer 6

final answer is:
http://www.wolframalpha.com/input/?i=%28-1151570.113-911120.2211i%29%5E%281%2F%28%28%285%E2%88%9A3%29%2F%282%29%29-%282.5i%29%29%29
theta used is -2.472241772 ---> i don't know where it came from
@ganeshie please help

Answer 7

how did u figure out theta used there is -2.4 ?
i am quite sure, the theta needs to be
\(\Large \theta = arctan(\dfrac{-911120}{-1151570})\)

Answer 8

yea i thought so too... but my classmate used 2.47... and his answer is the same as wolfram's

Answer 9

eh, just realized!

Answer 10

http://www.wolframalpha.com/input/?i=%E2%88%921151570.113%E2%88%92i911120.2211

Answer 11

\(\Large \theta = arctan(\dfrac{-911120}{-1151570})\)

gives you -2.4 in radians

Answer 12

since x and y are both negative,
the angle will be in 3rd quadrant = -141.6

which is -2.4 in radians

Answer 13

thank you!!

Answer 14

^_^