complex

[nicola de rosa]

Solve the following simple equation in $CC$:
$|z|*z^2+Re{z}*Im{z}-|z|^2*z=0$ with $z in CC$

[luca.barletta]

Consider the polar form $z=rhoe^(itheta)$.
Let's find the non-trivial solutions of the equation (trivial: $rho=0$):

$rhoe^(i2theta)+costhetasintheta-rhoe^(itheta)=0$ (1)

Immaginary part of (1):

$rhosin(2theta)-rhosintheta=0$

discarding the trivial solution:

$sin(2theta)-sintheta=0 rarr {(theta_1=kpi),(theta_2=pi/3+2kpi),(theta_3=-pi/3+2kpi):}, k in ZZ$

Real part of (1):

$rhocos(2theta)+sinthetacostheta-rhocostheta=0$ (2)

Make the substitution $theta=theta_1$ in (2):

$rho-rho(-1)^k=rho(1-(-1)^k)=0$

discarding the trivial solution, $k$ must be even, so $theta_1=2kpi, k in ZZ$.

Make the substitution $theta=theta_2$ in (2):

$-rho/2+1/2sqrt(3)/2-rho/2=0 rarr rho=sqrt(3)/4$

Make the substitution $theta=theta_3$ in (2):

$-rho/2-1/2sqrt(3)/2-rho/2=0 rarr rho=-sqrt(3)/4<0$

Finally, solutions are:

$(rho=0) vv (theta=2kpi, k in ZZ) vv (rho=+sqrt(3)/4, theta=+pi/3+2kpi, k in ZZ)$

[nicola de rosa]

My solutions are
$z=a$ $a>=0$ and
$z=sqrt(3)/4*e^(i*(pi/3+2kpi))$ $k in ZZ$

However $rho>=0$, so your solution $rho=-sqrt(3)/4$ isn't acceptable.

[luca.barletta]

Right, I edit

[nicola de rosa]

Solve in $CC$
$z^8=i*z^(**)*|z|$

[Camillo]

I will use the polar form ; consequently
$ z= rhoe^(itheta) $ ; $z^**=rhoe^(-itheta) $ ; $|z| =rho $ ; $i = e^(i*pi/2) $ and the equation becomes :

$rho^8 e^(i8theta) = rho^2e^i(pi/2-theta) $ which has the trivial solution $ rho = 0 $ and :
$rho ^6 e^(i8theta) = e^(i(pi/2-theta))$ .
Consequently : $ rho = 1 $ and $ 8theta = pi/2 -theta +2kpi $ and finally :
$ theta = (pi/18+2kpi/9)$ , $k in ZZ $.
Solutions are :
$ z = e^(i(pi/9)(1/2+2k)) $, $ k in ZZ $.