Question

\[{1+i}\over{-4-8i}\]

Answer 1

multiply and divide with -4+8i

Answer 2

what??

Answer 3

to make the denominator real,u need to multiply and divide by the complex conjugate of the denominator....u know whats complex conjugate of -4-8i ??

Answer 4

oh okay i remember now

Answer 5

then u can use the formula for denominator:
(a+bi)(a-bi)=a^2+b^2

Answer 6

what is i^2, again?

Answer 7

i^2 = -1

Answer 8

no.
(-4-8i)(-4+8i)=(-4)^2-(8i)^2=16-64i^2=16-64(-1)=16+64=80

Answer 9

\[{{12i+3}\over{80}}~~~~~?\]

Answer 10

the question says to write it in standard form....???

Answer 11

(i-3)/20

Answer 12

in denominator 80 should be there
in numerator you should have (i+1)(-4+8i)
after multiplication you should have.
-4i+8i^2-4+8i=-12+4i this is the numerator.

Answer 13

no its (1+i)(4+8i)

Answer 14

because it was (-4-8i) so the complex conj. is (4+8i)

Answer 15

nopes,for complex conjugate u flip the sign of imaginary part only!

Answer 16

the conjugate does not change the real part
so sonjugate should be -4+8i

Answer 17

oh okay right...so its\[{-12i+3}\over80\]

Answer 18

but how do i write that in standard form??

Answer 19

\[4i-12\over80\] is what im getting now

Answer 20

which is\[i-3\over20\]

Answer 21

in the numerator it should be -12+4i
standard form is a+bi
so
(-12+4i)/(80)
=-12/80+4i/80
=-3/20+1/20 i
here a=-3/20 and b=1/20
=-3/20+1/20 i is standard form.

Answer 22

(i-3)/20 is standard form?

Answer 23

you can say that. but usually complex no are expressed in
z=a+ib
if you expand it you will have
=-3/20+i/20

Answer 24

do u have choices,if not -3/20+i/20 is standard form....

Answer 25

thanks guys!

Answer 26

welcome :)