(3+4i)(3-4i)
just expand it as if the i was a variable
(3+4i)(3-4i)
= 9 - 12i + 12i -12i^2, now remember i^2 = -1
= 9 - 12(-1)
= 9+12
= 21
wait when you factor it wouldnt it be 9-12i+12i-16i^2
of course, my silly error
its totally cool you have helped alot :)
To multiply complex numbers, we can use the distributive property, just like with real numbers. The only difference is that we need to remember the special rule that \(i^2 = -1\). Here's how to multiply \( (3+4i)(3-4i) \):
Step 1: Multiply the real parts of the complex numbers:
\( 3 \times 3 = 9 \)
Step 2: Multiply the imaginary parts of the complex numbers:
\( 4i \times -4i = -16i^2 \)
Step 3: Simplify \( i^2 \) using the rule \( i^2 = -1 \):
\( -16i^2 = -16(-1) = 16 \)
Step 4: Combine the real and imaginary parts:
\( 9 + 16 = 25 \)
So, \( (3+4i)(3-4i) \) is equal to 25.