Question

(1-3i)(4+5i) @whpalmer4

Answer 1

use distributive multiplication
(1)(4) = 4
(1)(5i) = 5i
(-3i)(4) = -12i
(-3i)(5i) = -15i^2

Answer 2

I meant 4 +5i -12i -15i^2

Answer 3

which = 4 - 7i -15(-1) you know that i^2 = -1

Answer 4

then you have 4-7i+15
and that = -7i+19

Answer 5

\[(1-3i)(4+5i) = 1(4+5i)-3i(4+5i) = 1*4+1*5i-3i*4-3i*5i\]\[=4+5i-12i-15i^2\]Now substitute \(i^2=-1\) to give\[4+5i-12i-15(-1)=4+5i-12i+15=\](collect like terms to simplify)

Answer 6

19-7i