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

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You also may use the following alternative method that uses the property of absolute value, such that:

`|a/b| = |a|/|b|`

Considering `a = 3 - i` and `b = 3 - 2i` yields:

`|a| = sqrt(3^2 + (-1)^2) => |a| = sqrt(9+1) => |a| = sqrt 10`

`|b| = sqrt(3^2 + (-2)^2) => |b| = sqrt(9 + 4) => |b| = sqrt 13`

Replacing `sqrt 10` for `|a|` and `sqrt 13` for `|b|` yields:

`|(3-i)/(3-2i)| = sqrt 10/sqrt 13 => |(3 - i)/(3 - 2i)| = sqrt(10/13)`

Hence, evaluating the absolute value of fraction `(3 - i)/(3 - 2i) yields |(3-i)/(3-2i)| = sqrt(10/13).`

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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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