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

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### 4 Answers

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**

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**

(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`

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.**