How do you perform the operation and write the result in standard form given (2-3i)^2(2−3i)2?

2 Answers
Jim G.
Mar 26, 2018

-5-12i−5−12i

Explanation:

color(orange)"Reminder "color(white)(x)i^2=(sqrt(-1))^2=-1Reminder xi2=(√−1)2=−1

"note that "(2-3i)^2=(2-3i)(2-3i)note that (2−3i)2=(2−3i)(2−3i)

"expand factors using FOIL"expand factors using FOIL

=4-12i+9i^2=-5-12ilarrcolor(red)"in standard form"=4−12i+9i2=−5−12i←in standard form

kevincmiles1921
Mar 26, 2018

-5-12i−5−12i

Explanation:

First, multiply using the distributive property:

(2-3i)^{2} = (2-3i)(2-3i)(2−3i)2=(2−3i)(2−3i)

(2-3i)(2-3i) = 4 - 6i - 6i + 9i^{2}(2−3i)(2−3i)=4−6i−6i+9i2

Then simplify;

4 - 6i - 6i + 9i^{2} = 4 - 12i + 9(-1)4−6i−6i+9i2=4−12i+9(−1)

4 - 12i + 9(-1) = 4 - 12i -94−12i+9(−1)=4−12i−9

4 - 12i - 9 = -5-12i4−12i−9=−5−12i

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