Question

If $|a|=|b|$ and $\overline{a}c\ne b\overline{c}$, then the equation $az+b\overline{z}+c=0$ has

• A. No solution
• B. Exactly one solution
• C. Finitely many solutions
• D. Infinitely many solutions

Answer 1

Answer: (A)

No solution

Let $a=i$ and $b=1$ so that $|a|=|b|$.
$\overline{a}=-i$ and $\overline{b}=1$.
Let $c=p+iq$
Thus, $-ic\ne \overline{c}=>-ip+q\ne p-iq=>-i(p-q)\ne (p-q)=>p-q\ne 0=>p\ne q$.
Let $z=m+in$
$az+b\overline{z}+c=0=>iz+\overline{z}+p+iq=0=>im-n+m-in+p+iq=0=>(m-n+p)+i(m-n+q)=0$
Thus, $m-n+p=m-n+q=0=>p=q$.
But, we know that p and q are not equal. Hence, there is a contradiction.
Hence, (A) is correct.