How can complex roots and equations be solved using different methods?

[converting1]

Question: http://gyazo.com/abca582e1109884964913493487ad8ae

My solution:

I got √6 + i√2 = √8(cos(pi/6) + isin(pi/6)) as they did below,
then [itex] 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?

 

[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 $-\pi < \theta \leq \pi$

I got √6 + i√2 = √8(cos(pi/6) + isin(pi/6)) as they did below,
then $z^{\frac{3}{4}} = \sqrt{8}e^{i(\frac{\pi}{6} + 2k\pi)}$
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?

... 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.

 

[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.

 

[converting1]

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.

What I done:

$z^{\frac{3}{4}} = \sqrt{8}e^{i(\frac{\pi}{6})}$
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 = ...

 

[haruspex]

$z = (\sqrt{8}e^{i(\frac{\pi}{6} + 2k\pi)})^{\frac{4}{3}}$

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

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.

 

[converting1]

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.

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:
$$z = 4e^{i(\frac{2\pi}{9})}$$
$$z = 4e^{i(\frac{8\pi}{9})}$$
$$z = 4e^{i(\frac{-4\pi}{9})}$$

 

[haruspex]

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

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

 

[converting1]

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

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?

 

[haruspex]

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?

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

 

[converting1]

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

thank you for your patience! :)