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

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### 2 Answers

The absolute value of a complex number z= a + ib is given as |z| = sqrt (a^2 + b^2). The conjugate of a complex number z =a + ib is z' = a - ib.

Here we have 2z - z' = 3 + 2i

Let z = a + ib

=> 2a +2*i*b - a + i*b = 3 + 2i

=> a + 3*i*b = 3 + 2i

equate the real and complex parts

=> a = 3 and 3b = 2

=> a = 3 and b = 2/3

|z| = sqrt ( 3^2 + (2/3)^2)

=> sqrt ( 9 + 4/9)

=> sqrt (85/9)

=> (sqrt 85)/3

**The required absolute value is (sqrt 85)/3**

Given the expression:

2z ; z' = 3+ 2i

We need to find the absolute value of z.

First we will need to rewrite z using the form z =a+ bi

Then z' = a- bi.

Let us substitute.

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

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

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

==> a = 3

==> b= 2/3

==> z = 3 + 2/3 *i

Now we will calculate the absolute value.

We know that:

l zl = sqrt(a^2 + b^2)

==> l z l = sqrt(3^2 + (2/3)^2 = sqrt( 9 + 4/9) = sqrt(85/9)

**==> Then the absolute value for z is l z l = sqrt(85/9) = sqrt(85) / 3**