Determine the modulus of the complex number z=(3-i)/(3-2i)

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

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

First we need to rewrite the number by getting rid of the denominator.

We will multiply the numerator and denominator with (3+2i)

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

Let us calculate each:

(3-i)(3+2i) = 9 +6i - 3i - 2i^2  

We know that i^2 = -1

==> (3-i)(3+2i) = 9 + 3i + 2 = 11+ 3i

==> (3-2i)(3+2i)= 3^2 - (2i)^2

                        = 9 - 4i^2

                         = 9 + 4= 13

Now let us rewrite:

z= (11+ 3i)/13

   (11/13) + (3/13)i

Now to calculate the modulus:

the modulus is the absolute value of the number such that:

if z= a+ bi

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

==> l z l = sqrt(11/13)^2 + (3/13)^2

               = sqrt(121+ 9)/13^2 = sqrt130/ 13

==> l zl = sqrt130/ 13

 

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

Find the modulus of the complex number z = (3-i)/(3-2i).

We first convert the given complex number z = (3-i)/(3-2i) in the form x+iy.

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

We multiply both numerator and denominator of the right side by (3-2i) so that the denominator becomes real.

z = (3-i)(3-2i)/{(3-2i)(3+2i}

z = ( 9 - 6i -3i +2i^2)/(9 - (4i^2)) .

Z = (9-9i-2)/(9+4) , as i^2 = -1.

z = (7-9i)/13.

z = (7/13) + (-9/13)i.

|z| =sqrt{ (7/13)^2 + (-9/13)^2 } , |x+iy| = sqrt(x^2+y^2).

|z| = (1/13) sqrt{49+81}

|z| = (1/13) sqrt[130]

Therefore modulus of the given copmplex number = |z| = (1/13)sqrt(130).

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

The presence of the complex number from denominator is not allowed. According to the rule, we'll transform the complex number from denominator into a real number. For this reason, we'll multiply the ratio by the conjugate of the denominator.

The conjugate of denominator is 3+2i.

The ratio will become:

(3-i)/(3-2i) = (3-i)(3+2i)/(3-2i)(3+2i)

We'll notice that the denominator is a difference of squares:

(3-2i)(3+2i) = 9  - 4i^2

But i^2 = -1

(3-2i)(3+2i) =  9 + 4

(3-2i)(3+2i) = 13

We'll remove the brackets from numerator:

(3-i)(3+2i) = 9 + 6i - 3i - 2i^2

We'll combine like terms:

(3-i)(3+2i) = 9 + 3i + 2

(3-i)(3+2i) = 11 + 3i

The complex number z is:

z = (11 + 3i)/13

z = 11/13 + 3i/13

The modulus of the complex number is:

|z| = sqrt [Re(z)^2 + Im(z)^2]

|z| = sqrt[(11/13)^2 + (3/13)^2]

|z| = sqrt (121+9)/13

|z| = sqrt (130)/13

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