math expressionWrite the expression in the standard form (3-i)/(2+7i) .

2 Answers | Add Yours

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

In the standard form (3-i)/(2+7i) would have a real denominator.

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

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

=> [6 - 2i - 21i + 7i^2)/(4 - 49i^2)

=> (-23i - 1)/53

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

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

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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

The standard form is:

z = x + i*y

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 is:

z = -1/53 - 23i/53

 

The real part is Re(z) = -1/53, as it can be seen the row above.

The complex number is:

z = -1/53 - 23i/53

We’ve answered 318,911 questions. We can answer yours, too.

Ask a question