Question

if
\(\dfrac{(a+ib)^{x+iy}}{(a-ib)^{x-iy}} = p+iq\)

find, p, q
i don't have any answer for this.
Attaching my attempt

Answer 1

lol, I'll let the others check it for you this time. :)
they deserve it... xD

Answer 2

can easily separate real and imaginary parts
by will that do any good ?

Answer 3

Hahaha. My hunch here is that we should be multiplying by a complex conjugate.

Answer 4

complex conjugate of which varible, lol there are so many complex variables :P

Answer 5

I believe the next step is to FOIL

Answer 6

yes mira, foil, then separate real and imaginary parts, but will that help ?

Answer 7

Well my first reaction is to: \[\Large [\frac{(a+ib)}{(a-bi)} ]^x*[(a+ib)(a-ib)]^{iy}=p+iq\] I think it will allows us to remove some of our imaginary parts. Notice inside the fraction raised to the x power we can multiply by the complex conjugate of the denominator.

Answer 8

i believe so, one sec.

Answer 9

Actually I realized we might have a much simpler way of solving this if we convert to polar form: \[\Large a+bi = re^{i \theta}\] Then this means the complex conjugate is just: \[\Large a-bi = re^{-i \theta}\]

Answer 10

good idea!

Answer 11

It's definitely not pretty looking but it looks to be solvable haha.

Answer 12

it looks like the posted solution is a partial one and leads to another complicated set of equations.

Answer 13

ofcourse, it was just an attempt...

Answer 14

\[\Large \frac{(re^{i \theta})^{x+iy}}{(re^{-i \theta})^{x-iy}}=r^{2iy}e^{2i \theta x}e^{-2 \theta y}=e^{2i (\theta x+ y \ln r)}e^{-2 \theta y}\]\[e^{2i (\theta x+ y \ln r)}e^{-2 \theta y}=e^{-2 \theta y}\cos(2(\theta x+ y \ln r))+i e^{-2 \theta y}\sin(2(\theta x+ y \ln r))=p+iq\]

Now we've got two parts to equate... lol...

Answer 15

^ :O

Answer 16

looks complicated ;-;

Answer 17

these types of problems can lead to very involved non-linear coupled equations.

Answer 18

indeed

Answer 19

I'll leave that as an exercise to the reader... uhh...

Yeah actually you're right @Miracrown this isn't looking too good actually.

Answer 20

I would be very surprised if p and q here can be expressed in simple form... lol

Answer 21

Maybe there's really a beautiful answer to this we're not seeing? I always just hope that because I'd rather not spend time meticulously cranking through calculations...

Answer 22

I had started this problem by writing a + ib in polar form as well...

Answer 23

It could be the case that some very magical results will unfold here.

Answer 24

Actually I just realized that I have the answer. We simply need to just plug in theta and r from our original substitution of \[\LARGE a+bi = re^{i \theta}\] to get it back in terms of our equation.

I thought we were looking for a and b, but really we just need p and q, which is exactly what I have there.

Answer 25

we could use polar forms here

Answer 26

The solution presented in the Word attachment is a nice one. @hartnn
But since it doesn't run to completion, I have the suspicion that the missing parts will lead to another nasty set of coupled equations.

Answer 27

kainui, can't those be simplified....p,q will be very huge functions in a,b,x,y...

Answer 28

You can hope that upon plugging in: \[\LARGE r=\sqrt{a^2+b^2} \\ \LARGE \theta = \tan^{-1}(\frac{b}{a})\] simplifies it. But honestly I don't know since this is where I stopped since the rest is downhill from here and kind of ugly haha...

Answer 29

Yes, apparently this is the case.

Answer 30

until something easier comes up, i am gonna go with kainui's solution.

Answer 31

I'm playing around with doing the complex conjugate on everything in like 4 ways to see if I get something prettier out of it.

Answer 32

Eh, I doubt it. http://www.wolframalpha.com/input/?i=Re%7B(a%2Bbi)%5E%7Bx%2Biy%7D%2F(a-bi)%5E%7Bx-iy%7D%7D&t=crmtb01

Answer 33

I'll try to separately work this as well to see if anything decent comes from it.

Answer 34

I see that even Wolfram has its hands full with this expression. lol

Answer 35

I'll play with this problem over the next several minutes to see if it leads anywhere useful.

Answer 36

me too

Answer 37

Where did you get this problem from? Do the answers from this source generally have a clever way to get a beautiful result?

Answer 38

asked in university exam a year ago, they don't provide solutions

Answer 39

Hmm. It feels like this sort of puzzle:\[\LARGE LRIENAEDS\] It sort of points itself towards the right answer once you get it, READ between the LINES.

I just am not seeing how this one works, seems like a crappy test question. Hmm.

Answer 40

\[\large e^{\log \dfrac{(a+ib)^{x+iy}}{(a-ib)^{x-iy}}}\]


\[\large e^{(x+iy)\log(a+ib) - (x-iy)\log(a-ib)}\]

Answer 41

\[\large e^{(x+iy)(\log r + i\theta ) - (x-iy)(\log r - i\theta )}\]

Answer 42

hmm... @ganeshie have you been sleeping all this time ?

Answer 43

kindof.. :)

Answer 44

\(\huge e^{\Large 2iy \log r +2ix \theta }\)

just another way of what kainui did...

Answer 45

yes didn't expect it to simplify this nicely lol

Answer 46

\(e^{-2\theta y}\)
we don't get this term, but kainui's method had it... ?

Answer 47

i just copy pasted kainui's method for my own understanding :) both are same... look again

Answer 48

i think kainui did an error, y theta is actually getting cancelled

Answer 49

both numerator and denominator has
\(\Large e^{-i\theta y}\)

Answer 50

I don't think so, the exponent's both have i's so it shouldn't be an imaginary power.

Answer 51

\(\Large \frac{(re^{i \theta})^{x+iy}}{(re^{-i \theta})^{x-iy}}=r^{2iy}e^{2i \theta x}=e^{2i (\theta x+ y \ln r)}\)

Answer 52

i mean both num and denom has
\(e^{-\theta y}\)

Answer 53

isn't it ?

Answer 54

Oh so it is! I messed that one up.

Answer 55

now, not so complicated after all :D

Answer 56

\[\LARGE (r^ye^{x \theta})^{2i}\] is much nicer lol.

Answer 57

yeah, i pursued the problem using the solution method you attached.
There are some nice cancellations along the way. ^

Answer 58

there are ?! :O