How would you simplify 7i/(2-3i)?

1 Answer
Nov 2, 2017

-21/13+14/13i2113+1413i

Explanation:

To simplify a number with a complex denominator, we need to utilize the complex conjugate. For any complex number a+bia+bi, its complex conjugate, bar (a+bi)¯¯¯¯¯¯¯¯¯¯a+bi, is the complex number a-biabi. The conjugate is remarkably handy in turning complex denominators into real ones, which, in this case, will allow us to break apart our fraction into a real and imaginary part of the form a+bia+bi.

When we multiply a+bia+bi by bar (a+bi)¯¯¯¯¯¯¯¯¯¯a+bi, we can remember the fact that the particular case of the binomials (a+b)(a-b)(a+b)(ab) has the product a^2-b^2a2b2. Since ii has the special property that i^2=-1i2=1, (a+bi)(a-bi)(a+bi)(abi) actually ends up giving us a^2+b^2a2+b2; this is especially convenient, as the imaginary part is completely eliminated in the product.

Turning out attention back to the problem, we'd like to turn the denominator into a real number, but we can't change the value of our number as a whole, so we'll have to multiply it by some form of 1. To get our denominator 2-3i23i into the right form, we'll choose the fraction bar(2-3i)/bar(2-3i)¯¯¯¯¯¯¯¯¯¯23i¯¯¯¯¯¯¯¯¯¯23i, or (2+3i)/(2+3i)2+3i2+3i.

Multiplying that out, we get:

(7i)/(2-3i)*(2+3i)/(2+3i)=(7i(2+3i))/((2-3i)(2+3i))7i23i2+3i2+3i=7i(2+3i)(23i)(2+3i)

The denominator, as mentioned before, becomes 2^2+3^2=1322+32=13, and distributing the 7i7i to the 22 and the 3i3i in the numerator gets us 14i-2114i21. Altogether, we have

(14i-21)/1314i2113

which we can break apart into

(14i)/13-21/1314i132113

and rearrange into the form a+bia+bi:

-21/13+14/13i2113+1413i