# 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! :)