Find the absolute values of the complex number z = (2-3i)/(-2-4i)

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hala718's profile pic

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

Posted on

z = (2-3i)/ (-2-4i)

First we need to rewrite the complex number z in the form a+bi

Then we need to multiply and divide by the inverse of the denominator:

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

Let us simplify the numerator:

==> (2-3i)(-2+4i) = -4 + 8i + 6i - 12i^2

We know that i^2 = -1

==> (2-3i)(-2+4i) = -4 + 14i + 12

                            = 8 + 14i

Now for the denominator:

(-2-4i)(-2+4i) = (-2^2) - (4i)^2

                      = 4 - 16i^2

                      = 4 + 16 = 20

==> (-2-4i)(-2+4i) = 20

Now let us subsitute:

==> z = (8+14i)/20

             = 8/20  + (14/20)i

             = (2/5) + (7/10) i

Now let us calculate the absolute values:

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

        = sqrt[(2/5)^2 + (7/10^2 ]

         = sqrt(4/25 + 49/100)

          = sqrt(16+49)/100

           = sqrt(65) / 10

==> l z l = sqrt(65) / 10

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

To calculate the modulus of the complex number z, we'll re-write the given complex number z.

Since it is not allowed to keep a complex number to denominator, we'll multiply the ratio by the conjugate of the denominator, -2 + 4i:

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

We've obtained to denominator the difference of squares a^2 - b^2, where a = -2 and b = 4i => a^2 = 4 and b^2 = 16i^2.

But i^2  =-1 => b^2 = -16

a^2 - b^2 = 4 - (-16) = 4+16 = 20

z = (2 - 3i)(-2 + 4i)/(20)

 We'll calculate the numerator:

(2 - 3i)(-2 + 4i) = -4 + 8i + 6i + 12

We'll combine real parts and imaginary parts:

(2 - 3i)(-2 + 4i) = 8 + 14i

z = 2(4 + 7i)/20

z = (4 + 7i)/10

The modulus of z is:

|z| = sqrt (x^2 + y^2)

We'll identify x and y from the numbar z:

z = 4/10 + 7i/10

z = 2/5 + 7i/10

x = 2/5 and y = 7/10

|z| = sqrt (16/100 + 49/100)

|z| = sqrt (65/100)

|z| = [sqrt (65)]/10

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