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 + 14i^{2}
Now, it is where the small difference. It number is NOT in standard form. The standard form for complex numbers does not have an i^{2} in it. However, it is not a problem provided we recall that
i^{2} = -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 i^{2}.
(1 - 5i ) ( -9 + 2i ) = -9 + 2i + 45i -10i^{2} = -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 ) = a^{2} - abi + abi - b^{2}i^{2} = a^{2} + b^{2}
Thus, when we multiply a complex number by its conjugate we get a real number given by,
( a + bi ) ( a - bi ) =a^{2} + b^{2}