Complex roots and equations

1. May 15, 2013

converting1

My solution:

I got √6 + i√2 = √8(cos(pi/6) + isin(pi/6)) as they did below,
then $z^{\frac{3}{4}} = \sqrt{8}e^{i(\frac{\pi}{6} + 2k\pi)} [/tex] then took 4/3 of both sides and let k = 1, 0 etc to try and get the values of z however in the solutions: http://gyazo.com/3ba4f14d27f2fbc1bc8f04d2b918a5ac they raised it to 1/4 first then added the 2kpi and hence got different solutions than me. Could anyone explain why? 2. May 15, 2013 Simon Bridge The question is: Solve the equation $z^{3/4}=\sqrt{6}+\sqrt{2}i$ Put your answers in the form $z=re^{i\theta}$ where $r>0$ and $-\pi < \theta \leq \pi$ ... looking at the pic, it looks to me like they just raised both sides to the power of 4 to get the expression for $z^3$ ... if you take the cube root of both sides, it should be the same as your answer. 3. May 15, 2013 haruspex Your upload of the other solution is too hard to read. RHS is cropped. As far as I can tell, they also raised both sides to 4/3, effectively. Pls post your working and answer. 4. May 16, 2013 converting1 What I done: [itex] z^{\frac{3}{4}} = \sqrt{8}e^{i(\frac{\pi}{6})}$
as solutions are not unique:
$z^{\frac{3}{4}} = \sqrt{8}e^{i(\frac{\pi}{6} + 2k\pi)}$
$z = (\sqrt{8}e^{i(\frac{\pi}{6} + 2k\pi)})^{\frac{4}{3}}$

now I subbed in k = 0, k = 1... to get the solutions where -π<θ≤pi

what they have done
$z^{\frac{3}{4}} = \sqrt{8}e^{i(\frac{\pi}{6})}$
$z^3 = ( \sqrt{8}e^{i(\frac{\pi}{6})})^{4}$
$z^3 = 64e^{i(\frac{\pi}{6} + 2\pi k)}$
so $z = 4e^{i(\frac{2\pi}{9} + \frac{2k\pi}{3})}$

then they sub in k = 0, k = 1 and k = -1 to get values for z (only one of which is the same to mine).
I'm asking why do you have to add the 2pki after you get it into z^3 = ...

5. May 16, 2013

haruspex

Ok, but what are your solutions? The last line above is still consistent with their solutions. Following on from it:
$z = 4(e^{i(\frac{2\pi}{9} + \frac{8k\pi}3)})$
so the difference is a factor $e^{i\frac{6k\pi}3}$ = 1.
I agree it would be more natural to introduce the k term straight away, but I'm not seeing that deferring until after multiplying by 4 makes a difference. The important thing is to do it before dividing.

6. May 16, 2013

converting1

My only solution is $$z = 4e^{i(\frac{2\pi}{9})}$$, as all I can sub in is k = 0

they have three solutions of:
$$z = 4e^{i(\frac{2\pi}{9})}$$
$$z = 4e^{i(\frac{8\pi}{9})}$$
$$z = 4e^{i(\frac{-4\pi}{9})}$$

7. May 16, 2013

haruspex

Really? What do you get if you set k = 1?

8. May 17, 2013

converting1

then you get theta as 26pi/9... but this is out of the domain given for theta.

edit: hm, but as theta is not unique, is it OK to just take away 2pi from 26pi/9?

9. May 17, 2013

haruspex

Exactly so. You should reduce all results mod 2pi to get them into the target range.

10. May 18, 2013

converting1

thank you for your patience! :)