What is the standard form of the complex number ( 6 - 2i )/( 4 +14i )?

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The standard form of (6 - 2i )/( 4 +14i ) has a real denominator.

(6 - 2i )/( 4 +14i )

multiply numerator and denominator by (4 - 14i)

=> (6 - 2i )(4 - 14i)/( 4 +14i )(4 - 14i)

=> (24 - 8i - 84i + 28i^2)/(4^2 - 14^2*i^2)

=> (24 - 28 - 8i - 84i)/(16 + 196)

=> (-4 - 92i)/ 212

=> -1/53 - i*(23/53)

The simplified form is -1/53 - i*(23/53)

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giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

The standard form of the complex number is represented by the rectangular form, that  is:

z = x + i*y

The standard form does not allow for a complex numbers to be found at the denominator.

For the beginning, we'll factorize both, numerator and denominator, by 2:

( 6 - 2i )/( 4 +14i ) = 2(3 - i)/2(2 + 7i)

We'll simplify and we'll get: (3 - i)/(2 + 7i)

We'll have to get the complex number out of the denominator. For this reason, we'll have to multiply the numerator and denominator by the conjugate of the denominator: (2 - 7i).

(3-i)/(2+7i) = (3-i)*(2 - 7i)/(2+7i)*(2 - 7i)

(3-i)*(2 - 7i)/(2+7i)*(2 - 7i) = (6 - 21i - 2i + 7i^2)/(4 + 49), i^2 = -1

(6 - 21i - 2i - 7)/(4 + 49) = (-1 - 23i)/(53)

The standard form of the given complex number is:

z = -1/53 - (23/53)*i

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