# Find the absolute value of z is z= (2i-1)/(3+4i).

We have z= (2i-1)/(3+4i).

Now z= (2i-1)/(3+4i)

=> z = ( 2i - 1)*(3 - 4i) / (3 + 4i)(3 - 4i)

=> z = (6i - 3 - 8i^2 + 4i) / (3^2 - (4i)^2)

=> z = (10i - 3 + 8) / (9 + 16)

=> z = (10i + 5)/25

=> z = (10/25)i + (5/25)

Now |z| = sqrt ( (10/25)^2 + (5/25)^2)

=> |z| = sqrt ( 125) / 625)

=> |z| = sqrt 5/ 25

=> |z| = 1/ sqrt 5

Therefore the absolute value of z is 1/ sqrt 5.

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Given the complex number z = (2i-1)/(3+4i)

We need to find the absolute values of z = l z l.

First we need to simplify z and rewrite into the standard form z= a+ bi.

First we will multiply both numerator and denominator by (3-4i).

==> z = (-1+2i)(3-4i)/(3+4i)(3-4i)

= (-3 +4i +6i -8i^2)/(9 -16i^2)

But we know that i^2 = -1

==> z = ( -3+10i +8)/(9+16)

= ( 5+10i) / 25

==> z = (1/5) + (2/5)i

Now we will calculate the absolute values.

We know that l zl = sqrt(a^2+b^2)

==> l z l = sqrt(1/5)^2 + (2/5)^2

= sqrt(1+4)/25 = sqrt5 / 5

==> l z l = sqrt5 / 5 = 1/sqrt5

Approved by eNotes Editorial Team