Question

someone please help me :(

what is the quotient of the complex numbers below?
3+2i/1-5i
A. -24 -10i
B. -7/26 +17/26i
C. 3-2/5i
D. 13-13i

Answer 1

multiply top and bottom be the conjugate of \(1-5i\)

Answer 2

the conjugate of \(a+bi\) is \(a-bi\) and this works because \[(a+bi)(a-bi)=a^2+b^2\] a real number

Answer 3

all the work is multiplyiing in the numerator \[\frac{3+2i}{1-5i}\times \frac{1+5i}{1+5i}\]
\[=\frac{(3+2i)(1+5i)}{1^2+5^2}\]

Answer 4

you know how to multiply complex numbers?

Answer 5

no i don't, sadly.

Answer 6

you have a choice

Answer 7

you can just multiply out like you would say \((3+2x)(1+5x)\) you know how to do that?

Answer 8

lol it is not that sad, you can learn it
guess that is the point of this problem, but you can't divide if you don't know how to multiply for sure!

Answer 9

can you multiply this\[(3+2x)(1+5x)\]

Answer 10

you would combine like terms ? wouldn't you ?

Answer 11

you ever had a math teacher that said "foil" or sommat?

Answer 12

oh yes i just learned that, so 10x^2 + 17x + 3?

Answer 13

ok yes
so if you do it with \(i\) instead of \(x\) you get \[10i^2+17i+3\]but there is one difference

Answer 14

since \(i=\sqrt{-1}\) you know \(i^2=-1\) so \[10i^2+17i+3=-10+17i+3=-7+17i\]

Answer 15

and don't forget your denominator, which is \(1^2+5^2=1+25=26\)

Answer 16

giving you \[\frac{-7+17i}{26}\] or \[\frac{-7}{26}+\frac{17}{26}i\]

Answer 17

ok ok but i got so lost on that last step :(

Answer 18

the \(-1\) part?

Answer 19

yes

Answer 20

it is the definition of the symbol (letter) \(i\) used here
it is the number whose square is \(-1\) i.e. by definition \(i^2=-1\)

Answer 21

that is why \[10i^2+17i+3\] becomes \[-10+17i+3\]

Answer 22

oh so only i^2 = -1, not i ?

Answer 23

no the \(i\) stays there

Answer 24

if \(i\) was \(-1\) no one would write \(i\) they would just write \(-1\) instead

Answer 25

ok ok i get it now (: thank you !!! very helpful

Answer 26

yw