Example of multiplication of complex numbers, Mathematics

Multiply following and write the answers in standard form.

 (a) 7i ( -5 + 2i )

 (b) (1 - 5i ) ( -9 + 2i )

Solution

(a) Thus all that we have to do is distribute the 7i through the parenthesis.

7i (-5 + 2i ) = -35i + 14i2

Now, it is where the small difference. It number is NOT in standard form.  The standard form for complex numbers does not have an i2 in it. However, it is not a problem provided we recall that

                                                   i2  = -1

By using this we get,

 

7i ( -5 + 2i ) = -35i + 14 ( -1) = -14 - 35i

 We also rearranged order so that first the real part is listed.

 (b) In this we will FOIL the two of numbers and we'll have to also remember to get rid of the i2.

(1 - 5i ) ( -9 + 2i ) = -9 + 2i + 45i -10i2  = -9 + 47i -10 ( -1) = 1 + 47i

If we multiplied a number by its conjugate. There is a general formula for this which will be convenient while it comes to discussion division of complex numbers.

(a + bi ) (a - bi ) = a2 - abi + abi - b2i2  = a2 + b2

 

Thus, when we multiply a complex number by its conjugate we get a real number given by,

                                        ( a + bi ) ( a - bi ) =a2 + b2

Posted Date: 4/6/2013 3:23:18 AM | Location : United States







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