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

Mary Buchanan

Mary Buchanan

Answered

2022-01-18

How can we find z when the complex number Z satisfies the equation 4z3z=(18i2i),where z is the complex conjugate of z?

Answer & Explanation

einfachmoipf

einfachmoipf

Expert

2022-01-19Added 32 answers

The question would make more sense if it is about
4z+3z=18i2i
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:
18i2i=(18i)(2+i)4+1=216i+i8i25=23i
Then the equation implies, taking conjugates
{4z+3z=23i4z+3z=2+3i
Summing up we obtain 7(z+z)=4. Subtracting we get zz=6i. Therefore
z+z=47,zz=6i
and finally 2z=476i
Ian Adams

Ian Adams

Expert

2022-01-20Added 160 answers

Let Z=x+iy
4(x+iy)3(xiy)=18i2i12(x2+y2)=(18i)(2+i)(2i)(2+i)=2+i16i+84+112(x2+y2)=1015i5=23i
Left hand side is real. It can not be equal to a complex number. Hence there is no z satisfying the given condition.

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