Mary Buchanan

2022-01-18

How can we find z when the complex number Z satisfies the equation $4z3\stackrel{―}{z}=\left(\frac{1-8i}{2-i}\right)$,where $\stackrel{―}{z}$ is the complex conjugate of z?

einfachmoipf

Expert

The question would make more sense if it is about
$4z+3\stackrel{―}{z}=\frac{1-8i}{2-i}$
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:
$\frac{1-8i}{2-i}=\frac{\left(1-8i\right)\left(2+i\right)}{4+1}=\frac{2-16i+i-8{i}^{2}}{5}=2-3i$
Then the equation implies, taking conjugates
$\left\{\begin{array}{l}4z+3\stackrel{―}{z}=2-3i\\ 4\stackrel{―}{z}+3z=2+3i\end{array}$
Summing up we obtain $7\left(z+\stackrel{―}{z}\right)=4$. Subtracting we get $z-\stackrel{―}{z}=-6i$. Therefore
$z+\stackrel{―}{z}=\frac{4}{7},z-\stackrel{―}{z}=-6i$
and finally $2z=\frac{4}{7}-6i$

Expert

Let $Z=x+iy$
$\therefore 4\left(x+iy\right)3\left(x-iy\right)=\frac{1-8i}{2-i}\phantom{\rule{0ex}{0ex}}⇒12\left({x}^{2}+{y}^{2}\right)=\frac{\left(1-8i\right)\left(2+i\right)}{\left(2-i\right)\left(2+i\right)}=\frac{2+i-16i+8}{4+1}\phantom{\rule{0ex}{0ex}}⇒12\left({x}^{2}+{y}^{2}\right)=\frac{10-15i}{5}=2-3i$
Left hand side is real. It can not be equal to a complex number. Hence there is no z satisfying the given condition.

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