How can we find z when the complex number Z satisfies the equation 4z3z―=(1−8i2−i),where z― is the complex conjugate of z?

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einfachmoipf · Accepted Answer

The question would make more sense if it is about  4z+3z―=1−8i2−i  With the product in the left-hand side the equation would obviously have no solution, because the left-hand side is real and the right-hand side isn’t.  Write the right-hand side in rectangular form:  1−8i2−i=(1−8i)(2+i)4+1=2−16i+i−8i25=2−3i  Then the equation implies, taking conjugates  {4z+3z―=2−3i4z―+3z=2+3i  Summing up we obtain 7(z+z―)=4. Subtracting we get z−z―=−6i. Therefore  z+z―=47,z−z―=−6i  and finally 2z=47−6i

Ian Adams · Accepted Answer

Let Z=x+iy ∴4(x+iy)3(x−iy)=1−8i2−i⇒12(x2+y2)=(1−8i)(2+i)(2−i)(2+i)=2+i−16i+84+1⇒12(x2+y2)=10−15i5=2−3i Left hand side is real. It can not be equal to a complex number. Hence there is no z satisfying the given condition.

How can we find z when the complex number Z

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