17Calculus Precalculus - Multiplying Complex Numbers
17Calculus
Multiplying Complex Numbers
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Multiplying works the same as with radicals. For example, let's multiply the same two numbers we used above, first with square root of 2 and then with complex numbers.
Example: \((1+3\sqrt{2}) (6+8\sqrt{2}) = 1(6+8\sqrt{2}) + 3\sqrt{2}(6+8\sqrt{2}) =\) \( 6+8\sqrt{2} +18\sqrt{2} +24(2) =\) \( 54+26\sqrt{2}\)
Notice we had \( (\sqrt{2})^2 = 2 \).
Now lets multiply complex numbers.
Example: \((1+3i) (6+8i) = 1(6+8i) + 3i(6+8i) =\) \( 6+8i +18i +24i^2 =\) \( 6+26i - 24 = -18+26i\)
Notice that when we got \(i^2\) and since \(i=\sqrt{-1}\), this gives us \(i^2 = (\sqrt{-1})^2 = -1\). So \(24i^2 = 24(-1)=-24\).
Before we go on, let's watch a video to give us more information.
Dr Chris Tisdell - Complex numbers are AWESOME [16min-46secs]
video by Dr Chris Tisdell |
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