3z -i = 2 - 4i

First we need to rewrite z as a complex number of the form :

z = a+ bi

First we will add i to both sides:

==> 3z = 2 - 4i + i

==> 3z = 2 - 3i

Now let us divide by 3:

==? z = (2-3i)/3

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

Now we will determine the absolute value for z:

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

= sqrt( 2/3)^2 + 1^2)

= sqrt( 4/9 + 1) = sqrt( 13/9)

= sqrt13 / 3

Then the absolute values of z is :

**l z l = sqrt13 /3**

1) 3z -i = 2 -4i

4i -i = 2-3z

3i = 2-3z

3i + 3z = 2

3( i+ z) = 2

i + z = 2/3

z = 2/3 -i

2) 3z -i

= -3z + i (change the sign)

= -3 (z+ i)

= -3 (2/3) (substitute the value of (i + z )

= -2

3) 3z -i = 2- 4i( substitute the value above for (3z -i)

-2 = 2- 4i

4i= 2+ 2

i = 1

4) 3z - 1 = 2 - 4(1)

3z -1 = -2

3z = -2 +1

z = - 1/3 * ( u can check back the answer by substituting z value and i value in the formula ) The easiest way i know

To find the modulus of z in the equation:3z-i = 2-4i.

We know that if z = x+yi , x and y are real, then absolute value of z = |z| = sqrt(x^2+y^2).

Therefore we solve the equation in z in the form z = x+yi.

3z-i = 2-4i.

3z = 2-4i+i.

3z = 2-3i.

z = (2-3i)/3 = (2/3)+(-3/3)i = (2/3) + (-1)i.which is in x+yi form.

So |z| = sqrt{(2/3)^2+(-1)^2} = sqrt{4/9+1} = sqrt(13/9).

Therefore the absolute value of z = sqrt(13/9).