SOLUTION: Express (15 - 16i)(2 - 3i)$ in standard form. The standard form of a complex number is the form a + bi, where a and b are real.

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Question 1209237: Express (15 - 16i)(2 - 3i)$ in standard form.
The standard form of a complex number is the form a + bi, where a and b are real.
Found 2 solutions by Edwin McCravy, math_tutor2020:
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
The dollar mark student posts again:
Express (15 - 16i)(2 - 3i)$ in standard form.
The standard form of a complex number is the form a + bi, where a and b are real.
Use FOIL
%2815%29%282%29%2B%2815%29%28-3i%29%2B%28-16i%29%282%29%2B%28-16i%29%28-3i%29

Multiply those, and combine like terms, 
Remember that i2 = -1, 
so substitute -1 for i2.
combine terms again if necessary.
write the term without an i first,
and write the term with an i second.

Edwin

Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

I'll provide a similar example.

Let's say we want to evaluate (7-5i)*(4+9i)

I'll use the box method.

Write the terms 7 and -5i along the left side
Write the terms 4 and 9i along the top row
49i
7
-5i


To fill the four inner cells, we multiply the headers.
Example: -5i*9i = -45i^2 = -45*(-1) = 45 is in the bottom right corner.
49i
72863i
-5i-20i45


Terms along the northwest diagonal are the real components.
So 28+45 = 73 is the real part of the computation.

Meanwhile, terms along the northeast diagonal are the imaginary components.
-20i+63i = 43i

We determine that
(7-5i)*(4+9i) = 73+43i
Verification using WolframAlpha

The general template for multiplying two complex numbers is
(a+bi)*(c+di) = (ac - bd) + (bc + ad)*i

Another approach you could use is the FOIL method.