Question

express in polar form r cisθ for:
a) -1-i
b) -3+4i (to 2 decimal places of a degree)
c) use your answers to parts a) and b) to simplify (-3+4i)/(-1-i)^2 in polar form.

Answer 1

first you need \[r=\sqrt{a^2+b^2}\] so for part a you get
\[r=\sqrt{2}\] and for part b you get \[r=5\]

Answer 2

then you need \[\theta\] which you get by solving
\[tan(\frac{b}{a})=\theta\]

Answer 3

are you using degree or radians for this?
if radians the first one is \[\frac{5\pi}{4}\] if degrees is it 225

Answer 4

if you plot -1-i you see you are in quadrant III which is why you get 225

Answer 5

for part b you get \[tan^{-1}(-\frac{4}{3})\] from a calculator then you have to add 180 to your answer. let me see if i have one

Answer 6

i get 126.78 to two places but you should check it

Answer 7

so answer to first one is
\[\sqrt{2}(cos(45)+isin(45))\] and second is
\[5(cos(126.78)+isin(126.78))\]

Answer 8

then to divide, you divide the r's and subtract the angles.

Answer 9

what does it say next to (45)? its all blurry. thanks for your help so far!

Answer 10

ok first of all i made a mistake. it is not 45 it should be 225

Answer 11

i don't know why i wrote 45 because earlier i said the angle is 225 degrees

Answer 12

it says cos(225)+isin(225)

Answer 13

i guess the latex is hard to read but may be clearer if you refresh the browser, that often works for me

Answer 14

oh thanks for the tip!

Answer 15

your next job is to square
\[(-1-i)\] which is actually very easy not in polar form, but i guess they want it in polar form so we do it

Answer 16

you do it by squaring r and doubling the angle so you get
\[(-1-i)^2=\sqrt{2}^2(cos(225\times 2)+isin(225\times2))\]
\[=2cos(450)+isin(450))\]
\[=2(cos(90)+isin(90))\]

Answer 17

the last equality because sine and cosine are periodic with period 360 so i just subtracted 360 from 450 to get 90

Answer 18

btw this problem is really stupid because \[(-1-i)^2\] is just 2i and dividing by 2i is easy. very annoying to put in polar form. but in any case we see that since cos(90)=0 and sin(90)=1 we get \[(-1-i)^2=2i\] using polar form

Answer 19

your last job is to divide. as i said dividing by 2i is easy but they want you to use polar form so you divide 5 by 2 and subtract 90 from 126.78 to get the answer of
\[\frac{5}{2}(cos(36.78)+isin(36.78))\]

Answer 20

now the snap way: \[\frac{-3+4i}{(-1-i)^2}=\frac{-3+4i}{2i}\]
\[=\frac{-3+4i}{2i}\times \frac{-i}{-i}=\frac{4+3i}{2}=2+\frac{3}{2}i\]\]

Answer 21

thank you that answer matches the answer in the book :)

Answer 22

good. hope you understood the steps and that dividing by 2i is much easier without converting to polar for. easy to divide!