Question

(x+jy)^(x-jy) ??

Answer 1

you've gone too far :P
have u made this question on your own ?? :O

Answer 2

\(\Large (re^{j\theta })^{re^{-j\theta}}\)
nothing much can be done

Answer 3

(0.3468223942 -j0.01960894887)^(2-3i)

Answer 4

convert to exponential form?

Answer 5

you can use calculators, right ?
then u can use wolf too
http://www.wolframalpha.com/input/?i=%280.3468223942+-+i*0.01960894887%29%5E%282-3i%29

Answer 6

calcu says math error! haha

Answer 7

lol
thank god we have wolf to our rescue :)

Answer 8

i can't use this wolf in exams.. :'( haha just for now.. :DD

Answer 9

in exam, you can try converting into
\(\Large (re^{j\theta })^{re^{-j\theta}} = r^{r}r^{e^{-j\theta}}\times e^{rj\theta}e^{e^{-j\theta}j\theta} \)

Answer 10

that seems even more complex

Answer 11

\(e^{-j\theta} = \cos \theta -i\sin \theta \)
prolly that might help...

Answer 12

you calculator can do
(real)^(complex) ?

Answer 13

if it can, it solves all your problems

Answer 14

like 3^(5+8i)

Answer 15

eh , why j is complex ?

Answer 16

converted:
\[(0.3473e ^{j9.8574x10^-4}))^{\sqrt{13}e ^{j0.0172}}\]
?

Answer 17

can it do
0.347^(2-3i)
?

Answer 18

i guess exponent is not required to be in the form r e^(it)
it can remain in x+iy form

Answer 19

\[(x+jy)^{x-jy} = r^{(x-jy)/2}\cdot e^{j\theta (x-jy)} =\cdots \]
see this for a more general formula
http://mathworld.wolfram.com/ComplexExponentiation.html

Answer 20

\(j\) for engineers
\(i\) for mathematicians

@Marki

Answer 21

:O lol
i from imaginary :P
why u guys changed that >.<

Answer 22

i for current

Answer 23

i guess engineers have a problem of imagining >.<

Answer 24

lol ok fair enough !

Answer 25

\[\large \begin{align}&(0.3468223942 -j0.01960894887)^{2-3i} \\~\\& = \left(0.35e^{-i3.24}\right)^{2-3i}\\~\\&= 0.35^{2-3i} e^{-i3.24(2-3i)}\\~\\ &= 0.35^{2-3i} e^{-9.72 - 6.48i} \\~\\
&= 0.35^{2} e^{-9.72} 0.35^{-3i} e^{-6.48i}\\~\\
&= 0.35^{2} e^{-9.72} e^{-3i \ln (0.35)} e^{-6.48i}\\~\\
&= 0.35^{2} e^{-9.72} e^{-i(3 \ln (0.35)+6.48)} \\~\\

&= \cdots
\end{align}\]