Complex numbers: Solve ##Z^2\bar{Z}=8i##

  • Thread starter DottZakapa
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  • #26
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We have ##a\cdot (a^2+b^2)=0## and ##a^2+b^2\neq 0.## That is the same as ##x\cdot 2= 0##. Just as if ##a^2+b^2## was ##2## and ##a## was ##x##. It isn't, but it is the same situation.
##a^2+b^2\neq 0## isn't this assuming that ##a## and ##b## are purely real which isn't implied in the question?
 
  • #27
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##a^2+b^2\neq 0## isn't this assuming that ##a## and ##b## are purely real which isn't implied in the question?
##a,b## are real, since ##z## is written as ##z=a+ib## before the calculation started:
I have tried to substitute ##z=a+ib## solve the conjugate and the square, then separate the real from the imaginary and put all in a system, but becomes too complicated.
Now if ##a=b=0## then ##z=0## which doesn't solve the equation ##z^2\bar{z}=8i .##
 

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