# What is the complex number z from identity 3z-18z'=12+i?

*print*Print*list*Cite

### 2 Answers

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**.