Question 1106840: zz*+3(z-z*)=13+12i
Found 2 solutions by rothauserc, Edwin McCravy:
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! let z = a + bi
:
I assume by z*, you mean the complex conjugate of z, then
:
z* = a - bi
:
zz* = (a + bi)(a - bi) = a^2 -abi +abi -(b^2)(i^2) = a^2 +b^2
:
Note that i^2 = -1
:
3(z-z*) = 3(a+bi - a-bi) = 3(2bi) = 6bi
:
a^2 +b^2 +6bi = (a^2+b^2) +6bi
:
b = 2
a = 3
:
z = 3 +2i
:
Answer by Edwin McCravy(20054)  (Show Source):
You can put this solution on YOUR website!
zz*+3(z-z*) = 13+12i
Substitute x+yi for z and x-yi for conjugate z*,
where x and y are real.
(x+yi)(x-yi)+3[(x+yi)-(x-yi)] = 13+12i
Simplify the left side:
x²-y²i²+3[x+yi-x+yi] = 13+12i
Simplify using the fact that i² = -1
x²-y²(-1)+3[2yi] = 13+12i
x²+y²+6yi = 13+12i
Set the sum of the two real-numbered terms on the
left side, x²+y² equal to the real term, 13, on
the right side:
x²+y² = 13 <--equation #1
Set the sum of the two imaginary term on the
left side, 6yi equal to the imaginary term, 12i, on
the right side:
6yi = 12i
Divide both sides by 6i
y = 2
Substitute 2 for y in
x²+y² = 13
x²+2² = 13
x²+4 = 13
x² = 9
x = ±3
We get two answers:
z = x+yi = ±3+2i
So the two answers are z = 3+2i and x = -3+2i
Edwin
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