Conditions for solutions to aZ + b·Z̄ + c = 0

[PcumP_Ravenclaw]

Dear all, can you please help me show why the 2nd scenario below has no solution...

Theorem 3.3.1 Suppose that a and b are not both zero. Then the equation $aZ + b\bar Z + c = 0$
has
a, b and c are complex numbers. Z is the unknown complex number.

(1) a unique solution if and only if $|a| != |b|$;
(2) no solution if and only if |a| = |b| and $b\bar c != c\bar a$;
(3) a line of solutions if and only if |a| = |b| and $b\bar c = c\bar a$.

Answer:

(1) a unique solution

In the equation $aZ + b\bar Z + c = 0$
a,b and c are complex numbers.

for |a| = |b|, say $a = f + ig$
then |a| = $\sqrt{f^2 + g^2}$

negative f and g will be equal to positive because of the square root

so b can be $f + ig, f - ig, -f + ig, -f -ig, g + if, g - if, -g + if, -g - if$

if |a| and |b| are equal i.e. when a and b are any of the complex numbers above then there may be repeated solutions for z.

if $z = z1 + iz2$ then for two different a and b in the above list z1 will same and z2 will be same.

(2) no solution

if |a| = |b| then ONLY repeated solutions will occur and a and b are only one of the eight complex numbers given, c is any complex number?

what does the condition $b\bar c != c\bar a$ do?(3) line of solutions
(3) what does the condition $b\bar c = c\bar a$ do?

 

[chiro]

Hey PcumP_Ravenclaw.

The easiest way to do this (since it is a linear problem) is just set a = x + yi, b = d + ei, c = o + pi and Z = q + ri and get the solutions for the real and imaginary parts to equal 0.

If you do this, then it will be easier to answer your own questions.