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

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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 +...

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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

 

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