Question

$$ if \arg{z^{1/3}}= 1/2 \arg{z^2+\bar{z}.z^{1/3}}, \ then \ find\ |z|$$

Answer 1

@kc_kennylau can you help me in this.... :)

Answer 2

\[\Large arg(\sqrt[3]z)=\frac12arg(z^2)+\overline{z}\times\sqrt[3]z\]

Answer 3

Is this what you mean

Answer 4

I have no idea how you would add an angle to a complex number

Answer 5

no it is $$\huge 1/2 \arg({z^{2}+\bar z .z^{1/3}})$$

Answer 6

oh

Answer 7

\[\Large\arg\sqrt[3]z=\frac12\arg(z^2+\bar z\times\sqrt[3]z)\]

Answer 8

So this is what you mean

Answer 9

yes..... :P i forgot to put the brackets in the question.. :)

Answer 10

Should I let z be \(r(\cos\theta+i\sin\theta)\)

Answer 11

I don't know. I am just numb to this action. Just let me a way out.. :)

Answer 12

hey, is the $$2arg(x)= arg(x^2)$$ is it so. If, yes , than i can proceed. :).. May be...:P

Answer 13

\[2\arg(z)=2\theta=\arg(z^2)\]

Answer 14

Yep

Answer 15

I will try the rest thank you.. :)

Answer 16

\[\frac13\theta=\frac12\arg[r(\cos2\theta+i\sin2\theta)+r(\cos-\frac23\theta+i\sin-\frac23\theta)]\]