Evaluate and write in standard form: (-3 + 2i)^2 - 3(3 - i)(-2 + 2i),

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

(-3 + 2i)^2 - 3(3-i)(-2+2i)

We need to write into the standard form z= a+ bi

Let us open the brackets.

==> (-3+2i)^2 = (-3)^2 +2*-3*2i + 4i^2

But i^2 = -1

==> (-3+2i)^2 = 9 - 12i - 4 = 5- 12i

(3-i)(-2+2i) = -6 + 6i + 2i -2i^2 = -6 + 8i +2 = -4 +8i

==> 5-12i - 3(-4+8i) = 5-12i + 12 - 24i = 17 - 36i

Then the answer is :

z= 17 - 36i

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to simplify (-3 + 2i)^2 - 3(3 - i)(-2 + 2i).

(-3 + 2i)^2 - 3(3 - i)(-2 + 2i)

open the brackets

=> 9 + 4i^2 - 12i - 3( -6 + 2i + 6i - 2i^2)

=> 9 - 4 - 12i + 18 - 6i - 18i - 6 = 17 - 36i

The simplified form is 17 - 36i

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