Question

Simplify
6/(1+3i)

i is an imaginary number (square root of -1)

Answer 1

Do you know how to find the "conjugate" of 1 + 3i ? It's pretty simple.

Answer 2

multiply the inverse of (1+3i)\[\frac{ 6 }{ (1+3i) }\times \frac{ (1-3i) }{ (1-3i) }\] becaue now the denominator is (1-9i^2) and the 3i-3i cancel, so you get \[\frac{ 6-18i }{ (1-9i^2) }\] \[\frac{ 6-18i }{ 1-9(-1) }\] because i^2 is equal to -1 so the answer is \[\frac{ 6-18i }{ 10 }\] now simplify \[\frac{ 3 }{ 5 }-\frac{ 9 }{ 5 }i\]

Answer 3

wait not inverse... conjugate

Answer 4

celooscope:
Read the TOS for the website. You shouldn't be working out entire solutions for people.

Answer 5

sorry