Question

|z - 1| = 7z - 13 solve this equation and check for extraneous solutions

Answer 1

Given: |z-1| = 7z - 13

Due to the absolute value function, this equation yields two separate equations:

+(z-1) = 7z - 13, and

-(z-1) = 7z - 13.

Simplifying the first, we get z - 1 = 7z - 13, or

-1 + 13 = 7z - z, or 12 = 6 z. Then z = 2.

Doing the same to the 2nd equation: -z + 1 = 7z - 13, or 14 = 8z, or z = 7/4.

We must check both results by subst. into |z-1| = 7z - 13:

If z = 2, we get 2 - 1 = 14 - 13, or 1 = 1. Thus, z = 2 is a solution.

If z = 7/4, we get 3/4 = 7(7/4) - 13, or 3/4 = 49/4 - 52. This is clearly false.

The (single) solution is thus z = 2

Answer 2

`|z - 1| = 7z - 13`
either:

`z - 1 = 7z - 13`
and in that case:

`- 1 + 13 = 7z - z`

`12 = 6z`

`z = 12 / 6 = 2`
or:

`z - 1 = - 7z + 13`
and in that case

`z + 7z = 13 + 1`

`8z = 14`

`z = 14 * 8 = 1.75`
So; the answer is: either 2 or 1.75