I'm stuck in trying to solve this complex equation
5|z|3+2+3(ˉz)6=0
where ˉz is the complex conjugate.
Here's my reasoning: using z=ρeiθ I would write
5ρ3+2+3ρ6e−i6θ=05ρ3+2+3ρ6(cos(6θ)−i⋅sin(6θ))=0
from where I would write the system
{5ρ3+2+3ρ6cos(6θ)=03ρ6sin(6θ)=0
But here I get an error, since, from the second equation, I would claim θ=kπ6 for k=0…5, but θ=0 means the solution is real and the above equation doesn't have real solutions…where am I mistaken?
Answer
Let w=¯z3. Then we have
5|w|+2+3w2=0
As you point out, this constrains w=k or w=ki for real k.
Case 1. w=k
3k2+5|k|+2=0
which yields no solutions since the left-hand-size is always positive.
Case 2. w=ki
−3k2+5|k|+2=0
which yields k=±2, so w=±2i.
The rest is left as an exercise.
No comments:
Post a Comment