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hala718 eNotes educator| Certified Educator

(1+i)(1-3i)

Let us expand brackets:

(1+i)(1-3i) = 1*1 + 1*-3i + i*1 + i*-3i

                  = 1 -3i + i -3i^2

But we know that i^2 = -1

==> (1+i)(1-3i) = 1 - 3i +i -3*-1

                         = 1-2i +3

                          = 4 - 2i

==> (1+i)(1-3i) = 4 - 2i

giorgiana1976 | Student

To calculate the number, we'll have to remove the brackets:

(1+i)(1-3i) = 1*1 - 1*3i + i*1 - i*3i

(1+i)(1-3i) = 1 - 3i + i - 3i^2

We know that i^2 = -1

(1+i)(1-3i) = 1 - 3i + i + 3

We'll combine like terms (the real parts and the imaginary parts):

(1+i)(1-3i) = 4 - 2i

The result of the product (1+i)(1-3i) is a complex number whose real part is a = 4 and imaginary part, b = -2.

neela | Student

To calculate(1+i)/(1-3i)

Soluton:

(1+i)/(1-3i) = (1+i)(1+3i)/ (1-3i)(1+3i)

= (1-3 + 4i)/(1- -9) as i^2 = -1

= (-2 +4i)/10

= 0.2+0.4i

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