Question

(1+i)/(1-i); solve

Answer 1

would it be -i?

Answer 2

The product of conjugates normally gives you the DIFFERENCE of squares.
Example:
\[(a+b)(a-b)=(a^2-b^2)\]

But with complex numbers, since you're squaring the imaginary unit "i", you end up with the SUM of squares instead!
Example:
\[(a+bi)(a-bi)=(a^2+b^2)\]

Answer 3

If you're still confused you can let me know c:

Answer 4

is it A. -i ?
B. 1
C. or i

Answer 5

It is.... none of those... :O

Answer 6

is it -1
?

Answer 7

Hmmm it should be +2.. that's not an option? Hmm lemme look again.. maybe I made a silly mistake somewhere.

Answer 8

does it make a difference if it says simplify instead

Answer 9

hmmm

Answer 10

because then it is probs -i

Answer 11

OH AHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH

Answer 12

your problem is division, I didn't see the division symbol.. sorry that text is REALLY hard to read.

Answer 13

\[\frac{ 1+i }{ 1-i }\]
so ummm, to rationalize this, we'll multiply the top and bottom by the CONJUGATE of the bottom. so multiply it by (1+i)/(1+i)

Answer 14

\[[1^2-(i)^2]=[1-(-1)]=(1+1)=2\]
The bottom will simplify nicely like this.

Answer 15

Top gives us this:
\[(1+i)(1+i)=1^2+2i+i^2\]

Answer 16

here.

Answer 17

\[1^2+2i+i^2=1+2i+(-1)=2i\]\[\large \frac{ 2i }{ 2 }=i\]

Answer 18

so D?

Answer 19

Dude try to put in some effort lol.... sheesh :\

Answer 20

just making sure

Answer 21

Yah D sounds right, I think :O