How do I rationalize imaginary denominators? Such as..[ 5i/-2-6i ]? I'm so lost in Algebra II.It tells me to simplify the problem?

1 Answer | Add Yours

embizze's profile pic

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

Posted on

Simplify `(5i)/(-2-6i)` :

You can look at this as a simplification, or as division by imaginary numbers. The process is the same.

The answer will be a complex number of the form a+bi with a and b real numbers. There will not be any imaginary numbers in the denominator.

To get rid of the imaginary number in the denominator, we multiply the fraction by a special form of 1; the complex conjugate of the denominator over itself. (If z=a+bi then the complex conjugate of z is `bar(z)=a-bi` ) This process is called rationalization -- a similar procedure is used to get rid of radicals or radical expressions from the denominator.

`(5i)/(-2-6i)*(-2+6i)/(-2+6i)`

`=(-10i+30i^2)/(4-36i^2)` but `i^2=-1` by definition so

`=(-30-10i)/(40)`

`=-3/4-1/4i` which is the simplified form we seek.

** Multiplying by the complex conjugate makes use of the difference of two squares form; `(a+b)(a-b)=a^2-b^2` . This eliminates the term with i, so the denominator is now real. **

We’ve answered 318,982 questions. We can answer yours, too.

Ask a question