Wednesday, 31 August 2016

Solve complex equation 5|z|3+2+3(barz)6=0



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+3ρ6ei6θ=05ρ3+2+3ρ6(cos(6θ)isin(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=05, 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.


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