Evaluate and write in standard form: (4 - 2i)(-3 + 3i).

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The standard form of (4 - 2i)(-3 + 3i) can be written as

(4 - 2i)(-3 + 3i)

open the brackets and multiply

=> 4*-3 + 4*3i + 6*i - 6*i^2

=> -12 + 12i + 6i + 6

=> -6 + 18i

The required result is -6 + 18i

hala718's profile pic

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

Posted on

Let us assume that z= (4-2i)(-3+3i)

 We need to rewrite into the form z= a+ bi.

Let us simplify by opening the brackets.

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

==> x = -12 + 18i - 6i^2

But i^2 = -1

==> z= -12 + 18i + 6

          = -6 + 18i

==> z = -6 + 18i.

Then the values of the number (4-2i)(-3+3i) = -6 + 18i

 

atyourservice's profile pic

atyourservice | Student, Grade 11 | (Level 3) Valedictorian

Posted on

(4 - 2i)(-3 + 3i)

-12+12i+6i-6i^2

if the i stands for an imaginary number than the answer is -6 + 18i

but if it is just a variable than the answer is 

`-6i^2+18i-12`

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

The standard form of the equation is:

z = a + bi

We'll remove the brackets using FOIL method:

(4 - 2i)(-3 + 3i) = -12 + 12i + 6i - 6i^2, but i^2 =-1

We'll group the real parts and imaginary parts:

(4 - 2i)(-3 + 3i) = -12 + 18i + 6

(4 - 2i)(-3 + 3i) = -6 + 18i

The standard form of the equation is: (4 - 2i)(-3 + 3i) = -6 + 18i.

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