Question

Identify the quotient in the form a + bi.
6i/3+2i

Please show your work!

Answer 1

`(12)/(13)+(18)/(13)i`

Explanation:

Multiply by the conjugate as a factor of one:

`(6i)/(3+2i)=(6i)/(3+2i)*(3-2i)/(3-2i)=(18i+12)/(9+4)=(12+18i)/(13)=(12)/(13)+(18)/(13)i`

Answer 2

Answer:
`(12)/(13)+(18)/(13)i`

The expression is in a+bi form where,

a = 12/13

b = 18/13

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Work Shown:

Multiply top and bottom by 3-2i, which is the conjugate of the original denominator. This will make the denominator go from a nonreal complex number to a purely real number.

`(6i)/(3+2i)\\\\\\(6i(3-2i))/((3+2i)(3-2i))\\\\\\(18i-12i^2)/((3)^2-(2i)^2)\\\\\\(18i-12i^2)/(9-4i^2)\\\\\\(18i-12(-1))/(9-4(-1))\\\\\\(18i+12)/(9+4)\\\\\\(12+18i)/(13)\\\\\\(12)/(13)+(18)/(13)i\\\\\\`

The last expression is in a+bi form where a = 12/13 and b = 18/13

Keep in mind that
`i = √(-1)` has both sides square to
`i^2 = -1`. Also, I'm using the difference of squares rule to go from
`(3+2i)(3-2i)` to
`(3)^2 - (2i)^2`.