The complex number z satisfies the equation z+2iz*=12+9i. Find z giving your answer in the form x+iy.
if z = a+bi,
a+bi + 2i(a-bi) = 12 + 9i
a+2b + (2a+b)i = 12 + 9i
a+2b=12
2a+b=9
a = 2
b = 5
z = 2+5i
so you do a simultaneous equation by equating the variables?
I don't quite understand the second line
you didn't expand the brakets
OOOOHHHHHH get it now!! thank you Steve!!
Nice when the light bulb turns on, isn't it?
To find the complex number z that satisfies the equation, we can use algebraic manipulation.
Given the equation z + 2iz* = 12 + 9i, we can substitute z* with the conjugate of z, which is z* = x - yi, where z = x + yi.
Substituting the values into the equation, we have:
z + 2i(x - yi) = 12 + 9i
z + 2ix - 2yi = 12 + 9i
Now, let's separate the real and imaginary terms:
(x + 2ix) + (-2y + 2iy) = 12 + 9i
The real parts and imaginary parts on both sides of the equation should be equal. Therefore, we can equate the real parts and the imaginary parts separately:
Real part: x + 2ix = 12 (Equation 1)
Imaginary part: -2y + 2iy = 9 (Equation 2)
Solving Equation 1 for x:
x(1 + 2i) = 12
x = 12 / (1 + 2i)
Now, let's simplify Equation 2 for y:
-2y + 2iy = 9
2iy - 2y = 9
2i(y - y) = 9
2i(0) = 9
0 = 9
This equation has no solution for y since 0 cannot equal to 9.
Therefore, there is no complex number z that satisfies the given equation.