# Find the absolute values of z if 2z-i +5 = 3z -3i +1

*print*Print*list*Cite

### 3 Answers

Given the complex equation:

2z -i + 5 = 3z -3i +1

We need to find the absolute value of z.

First we need to rewrite z into the format of the complex number z = a+bi

let us combine terms with z on the left side.

==> 2z -3z = -3i +1 +i -5

==> -z = -4 -2i

Now we will multiply by -1

==> z = 4 + 2i

Now we will calculate the absolute values.

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

= sqrt( 4^2 + 2^2)

= sqrt(16 + 4) = sqrt20 = 2sqrt5

**==> l zl = 2sqrt5**

We have to find the absolute value of z given that 2z-i +5 = 3z -3i +1

2z - i + 5 = 3z -3i +1

=> 3z - 2z = -i + 3i + 5 -1

=> z = 2i + 4

The absolute value of z is given by sqrt ( 2^2 + 4^2)

=> sqrt ( 4 + 16)

=> sqrt 20

**Therefore the absolute value of z is sqrt 20.**

### User Comments

We first solve the equation 2z-i +5 = 3z -3i +1 and then find the absolute value of z.

We subtract 3z from both sides:

2z-i+5-3z = = -3i+1.

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

We add i-5 to both sides:

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

-z = -2i-4.

We multiply both sides by -1.

z = 4+2i.

We know that the absolute vale of z = x+iy = sqrt(x^2+y^2).

Therefore the absolute value of z = (4+2i) is sqrt(4^2+2^2) = sqrt(16+4) = sqrt 20 = 2sqrt5.

Therefore the absolute value of z = 2sqrt5.