# What is the absolute values of z is 3z-1 = z -i + 5

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### 3 Answers

We have 3z - 1 = z - i + 5. We find z using this.

3z - 1 = z - i + 5

=> 3z - z = 1 + 5 - i

=> 2z = 6 - i

=> z = 3 - i/2

Now the absolute value of z is sqrt [ 3^2 + (1/2)^2]

= sqrt ( 9 + 1/4)

=> sqrt ( 37/4)

=> (sqrt 37) / 2

**The required absolute value is (sqrt 37) / 2**

Given 3z -1 = z -i + 5

To find the absolute value, first we need to rewrite z into the form of the complex number z = a + bi.

Let us isolate z on the left side.

==> 3z -1 = z -i + 5

==> 3z = z -i + 6

Now subtract z.

==> 2z = 6 - i

Now we will divide by 2.

==> z = 3 - (1/2)*i

Now that z is in the form a+bi we will calculate the absolute value.

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

==> l z l = sqrt(3^2 + (1/2)^2 = sqrt(9+1/4) = sqrt(37/4)

**==> l z l = sqrt37 / 2**

Q: What is the absolute value of z is 3z-1 = z -i + 5.

Solution:

If z = x+iy, then the absolute value of z = |z| = (x^2+y^2)^(1/2), where x and y are real.

We first solve for z from the given equation 3z-1 = z-i+5.

Subtract z-1:

3z-z = -i+5 +1

2z = 6-i which is x+iy form

So absilote z = |z| = {6^+ (-1)^2}^(1/2)

|z| = (36+1}^(1/2) = 37^(1/2).

**Therefore absolute value of z = 37^(1/2).**