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

Answer 1

Elijah Allen

#-5-12i#

#color(orange)"Reminder "color(white)(x)i^2=(sqrt(-1))^2=-1#

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

#"expand factors using FOIL"#

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

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Answer 2

Ariana Alvarez

#-5-12i#

First, multiply using the distributive property:

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

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

Then simplify;

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

#4 - 12i + 9(-1) = 4 - 12i -9#

#4 - 12i - 9 = -5-12i#

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Answer 3

Chloe Alexander

To perform the operation ( (2 - 3i)^2 ) and write the result in standard form, we first expand the expression using the binomial square formula ( (a - b)^2 = a^2 - 2ab + b^2 ), where ( a = 2 ) and ( b = 3i ).

[ (2 - 3i)^2 = (2)^2 - 2(2)(3i) + (3i)^2 ]

[ = 4 - 12i + 9i^2 ]

Since ( i^2 = -1 ), we substitute it into the expression:

[ = 4 - 12i + 9(-1) ]

[ = 4 - 12i - 9 ]

[ = -5 - 12i ]

So, the result of ( (2 - 3i)^2 ) in standard form is ( -5 - 12i ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.