Question

someone guide me through this problem. I dont know how to solve it. [(-1-i)^2 x (2i)^2]/(2-2i)

Answer 1

I'm guessing they want you to rationalize the denominator. So you will have to multiply top and bottom by 2+2i. This is the conjugate of 2-2i

Answer 2

Convert all complex numbers to trigonometric form and then simplify the expression. Write the answer in standard form.

Answer 3

oh gotcha

Answer 4

I tried to rationalize... I got a huge mess... thats why I am hoping someone can guide me through it

Answer 5

were you able to convert (-1-i)^2 to trig form?

Answer 6

well i didnt do that part yet bc of the number next to it

Answer 7

-1-i is in the form a+bi

a = -1
b = -1

theta = arctan(b/a) = arctan(-1/(-1)) = arctan(1) = 45 degrees

theta is in Q3, so add 180: 45+180 = 225 degrees

Answer 8

r = sqrt(a^2 + b^2)

r = sqrt((-1)^2 + (-1)^2)

r = sqrt(2)

Answer 9

therefore, -1-i converts to sqrt(2)*[cos(225) + i*sin(225)]

Answer 10

and you can use DeMoivre's theorem to find [sqrt(2)*[cos(225) + i*sin(225)]]^2

Answer 11

wait ... why would the entire formula be sqred? it is multiplying something else that needs to be sqred

Answer 12

well initially (-1-i) is squared

so because it is equivalent to sqrt(2)*[cos(225) + i*sin(225)], I'm squaring all of sqrt(2)*[cos(225) + i*sin(225)]

Answer 13

ok but what abt the 2isqrd

Answer 14

well (2i)^2 is simply 4i^2

you square 2 to get 4
you square i to get i^2

Answer 15

but I guess you could also convert that to trig form too

Answer 16

-4 but then what

Answer 17

2i = 0+2i

a = 0
b = 2

theta = arctan(b/a) = arctan(2/0) = undefined

so theta is 90 degrees

r = 2 because 2i is 2 units away from the origin

Answer 18

ok what do i do with that?

Answer 19

dont know what to do with 0+2i how would i put it into the form... of 2(cos90?

Answer 20

2i = 2*[cos(90) + i*sin(90)]

Answer 21

so far we have

-1-i = sqrt(2)*[cos(225) + i*sin(225)]

and

2i = 2*[cos(90) + i*sin(90)]

Answer 22

what do you get when you square sqrt(2)*[cos(225) + i*sin(225)]

Answer 23

2 cos90 + isin90

Answer 24

so now i multiply them??

Answer 25

both sets?

Answer 26

yes but after you square sqrt(2)*[cos(225) + i*sin(225)]

Answer 27

what did you get for that piece?

Answer 28

thats what i got... 2cos 90....

Answer 29

because it went to 450 but then i used the coterminal

Answer 30

oh gotcha, so

(-1-i)^2 = 2*[cos(90) + i*sin(90)]

I see what you meant now

Answer 31

if

2i = 2*[cos(90) + i*sin(90)]

then what is (2i)^2 equal to

Answer 32

what?

Answer 33

if you squared 2*[cos(90) + i*sin(90)], what do you get

Answer 34

oooo 4(cos180

Answer 35

so

(2i)^2 = 4*[cos(180) + i*sin(180)]

Answer 36

now you multiply 2*[cos(90) + i*sin(90)] and 4*[cos(180) + i*sin(180)]

Answer 37

do you know the shortcut to do so?

Answer 38

u mean just multiply? 8cis270?

Answer 39

yeah 2cis(90) with 4cis(180) to get 8cis(270), good

Answer 40

so all that work transforms each piece in the numerator to trig form, then you multiply to get 8cis(270)

Answer 41

now convert the denominator 2-2i to cis form

Answer 42

i got 2sqrt2 cis 45

Answer 43

2-2i is not in Q1, it's in Q4

Answer 44

so the angle isn't 45, but it is ???

Answer 45

2sqrt2cis 135

Answer 46

well you were on the right track with the 45, but it's really -45

-45 + 360 = 315

so 2-2i = 2*sqrt(2)*cis(315)

Answer 47

okok got it.. i was confused...

Answer 48

so up top, things simplified to 8*cis(270)

in the denominator, we have 2*sqrt(2)*cis(315)

Answer 49

got it

Answer 50

i figured it out.. (well mainly thanks to u) thank you so very much

Answer 51

what answer did you get

Answer 52

so it came otu to be 2-2i

Answer 53

yeah all of that and it turns out the answer is buried in plain sight: in the denominator

lol

Answer 54

yes I was thinking that... but its not always the case