# What is the absolute value of z if 2z- z' = 3+4i?

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### 2 Answers

Given the equation 2z -z' = 3+ 4i

We need to find the l zl

First we will rewrite z into the form a+ bi

==> z' = a- bi

Now we will substitute.

==> 2(a+bi) - (a-bi) = 3+ 4i

==> 2a + 2bi - a + bi = 3+ 4i

==> a + 3bi = 3+ 4i

==> a = 3

==> 3b = 4==> b = 4/3

Then z = 3 + (4/3)*i

Now we will calculate the absolute value.

==> l zl = sqrt(a^2 +b^2 ) = sqrt(3^2 + (4/3)^2

= sqrt(9 + 16/9) = sqrt( 81+16)/9 = sqrt97/9 = sqrt97 /3

**Then the absolute value of l z l = (sqrt97)/3**

To determine the absolute value of z, we'll have to determine first, the real part and the imaginary part of z.

z = a + bi

Re(z) = a and Im(z) = b

z' is the conjugate of z, therefore z = a+ bi and z' = a - bi

2(a+bi) - a + bi = 3 + 4i

2a + 2bi - a + bi= 3 + 4i

a + 3bi = 3 + 4i

Comparing both sides, we'll get: a = 3 and 3b = 4 => b = 4/3

The absolute value of z:

|z| = sqrt[Re(z)^2 + Im(z)^2]

|z| = sqrt[3^2 + (4/3)^2]

|z| = sqrt(9 + 16/9)

**The absolute value is: |z| = (sqrt97)/3**