Question

Numbers in standard form (a+bi)
(1-3i)/(2+i), the answer I got was (-2-7i), taking it to SF is -2/5-7i/5 is that the standard form or did I screw up?

Answer 1

@Jack1 if you could take a quick look ~(;W;)~

Answer 2

hey minato ;D

divided by complex numbers
(ur *i* here is sqrt -1, yeah? )

Answer 3

remember that 1/i = - i ...yeah?

Answer 4

Yush i=sqrt(-1)

Answer 5

Did not remember that ono

Answer 6

so to work it out, i find it's easier to times top and bottom to remove the i
so\[\frac {(1-3i)}{(2+i)} \rightarrow \frac {(1-3i) \times (2-i)}{(2+i) \times (2-i)}\]

so focus on the denominators here, pretty little trick:
FOIL it out like normal first:
(2+i) x (2-i) = 2x2 + 2 (-i) + 2i×2 +i(-i)
= 4 - (i^2) and i^2 = -1
so = 4 - (-1) = 5


so denominator turns to 5

so equation is now:
\[\frac {(1-3i) \times (2-i)}{5}\]

can u foil out the top line and see how u go from here dude?

Answer 7

sorry, doubled up there, should be
*(2+i) x (2-i) = 2x2 + 2 x (-i) + 2 x i +i(-i) [not 2i x 2)

Answer 8

(1-3i)(2-i) = 2-i-6i+4i^2 = 2-7i+4i^2= 2-4-7i=-2-7i
i^=-1

Answer 9

-2-7i over 5? Does that mean I was right?

Answer 10

do over:
(1- 3i ) (2 - i) = 1x2 + -i + -6i + 3i^2
= 2 -i -6i + 3(-1)
= 2 - 3 -7i
=> so = -1 - 7i

Answer 11

so that's numerator
and 5 is denominator
u can either fraction it up or decimal? up to u

Answer 12

God I am such a fish stick, I kept writing -3i (-i)=4i^2

Answer 13

haha, all good it too me 2 attempts 2, i forgot the 3 in the first go ;D

Answer 14

Sanka you Jack1-san, again. *0*

Answer 15

all good dude, anytime hey