What is the complex number z from identity 3z-18z'=12+i?
The complex number z to be determined satisfies: 3z - 18z' = 12 + i
Let z = a + ib , z' = a - ib
3z - 18z' = 12 + i
=> 3(a + ib) - 18(a - ib) = 12 + i
=> 3a + 3ib - 18a + 18ib = 12 + i
equate the real and imaginary coefficients
-15a = 12 and 21b = 1
=> a = -12/15 and b = 1/21
The complex number z = (-4/5) + (1/21)i
We'll consider z' as being the conjugate of z, therefore z = a + bi and z' = a - bi.
We'll replace z and z' by its rectangular forms, into the given identity:
3(a+bi) - 18(a - bi) = 12 + i
We'll remove the brackets:
3a + 3bi - 18a + 18bi = 12 + i
We'll combine real parts and imaginary parts form the left side:
-15a + 21bi = 12 + i
Comparing, we'll get:
-15a = 12 => a = 12/-15 => a = -4/5
21b = 1 => b = 1/21
The requesed complex number is: z = -4/5 + i/21.