Question

(6-i)/1+i ..how to simplify it . what rule applies here>??

Answer 1

\[ \large \frac{6-i}{1+i}\cdot\frac{1-i}{1-i}= \]

Answer 2

I did this step . then I am confused in the next step

Answer 3

Which part u confused?

Answer 4

multiply top and bottom by the conjugate of 1+i = 1 - i

Answer 5

just times it out like how u multiply quadratics

Answer 6

so the bottom is not 1-i^2?

Answer 7

(a+b)(a+b) << same as (1+i)(1-i)

Answer 8

yes

Answer 9

i^2 is equivalent to negative 1

Answer 10

so it will be 1 - (-1)

Answer 11

and what about the denumerator ? in the answer it is 2?

Answer 12

numerator is 5-7i

Answer 13

so 1-i and 1-i add?

Answer 14

yes

Answer 15

(6-i)(1-i)=
(6 - 6i -1 + i^2)

Answer 16

yes this is what I got . I am confused about the denumerator

Answer 17

how to multiply it?

Answer 18

denumerator...? u mean denominator?

Answer 19

yes

Answer 20

(1+i)(1-i)
(1 -i + i -i^2)

Answer 21

because i^2 = -1

Answer 22

so it will be (1 - (-1) ) which equals 2

Answer 23

ohh so I have to solve it too by rule

Answer 24

yea..

Answer 25

very help ful . its was not mentioned in the book .. Thanx

Answer 26

no problem xD by the way,
i= square root of -1
i^2 is just 1
i^3 = i
and so on...