polar form angles (1 Viewer)

[sasquatch]

When writing angles in the polar form is it best to..

e.g.

A = tan^-1(2/-2)
= tan^-1(-1)
= -pi/4

OR

tan A = 2/-2
-tan A = 1
tan(pi - A) = 1
A = pi - tan^-1(1)
= pi - pi/4
= 3pi/4

the answer says -pi/4 but im more used to writing positive angles. the domain given for the angle is -pi < A <= pi but both angles fit in the domain so which is best to use?

 

[richz]

both are rite, u can do wat u want

 

[sasquatch]

ok im confused..

could someone please tell me the angle ranges for each quadrant...

 

[richz]

u can use both ok??

here try this

(quad 2) | (quad 1)
sin +ve | All are +ve
180-@| @
---------------------------------
(quad 3) | (quad 4)
tan +ve | cos +ve
180+@| 360-@

to cut a long story short

All Stations to Cabra

 

[sasquatch]

no what i mean is from z = 2 -2 i

you can see that the point will be in the fourth quadrant

but because the angle range is -pi < A <= pi or -180 <= 180 then -tanA=(2pi - A) does not work.

instead i tired

-tanA = tan(-pi - A) but that doesnt work either.

wait nevermind...

 

[richz]

dont really understand wat ur talking about?

 

[sasquatch]

looky here..

with z = -1 + iroot3 for example

tan A = root3 / -1
tan A = -root 3

-tanA = root 3

and using the angle ranges of a circle you could say

tan(pi-A) = root 3 or tan(2pi - A) = root 3
A = pi - tan^-1(root3) or A = 2pi - tan^1(root3)
= 2pi / 3 or A = 5pi/3

if you notice 5pi/3 is out of the range -pi < A <= pi

even though it can be seen that the point is in the second quadrant because of the cordiantes (-1, root3), how can you solve for the angles using the standard relationships between angles within a circle.

i was just mucking around and as the angle ranges are shifted 180 degrees (or pi radians) it can be said that:

q1: tan A = tan(180 + A)
q2: -tan A = tan(-A)
q3: tan A = tan A
q4: -tan A = tan(180 - A)

is that valid as i just tested it out and seems to work and give the proper angles for the range specified by polar coordinates.

ok yeah so is it best to use something like above or finding the possible angles and subtracting 180 degrees or pi radians from the answer to give the angle in polar form because the booklet im working from does not explain this.

 

[sasquatch]

im having a hard time trying to do tese type of questions..could someone please tell me...even though i have the correct answer if my method is correct?


1)
z = -1 - i

|z| = Root(1 + 1)
= root2

tan A = -1/-1
=1

tan A = 1
A = pi/4

but -pi < A <= pi

therefore A = -3pi/4

therefore z = root2 cis -3pi/4


2)
z = 2 - 2i

|z| = root(4 + 4)
=2root2

tan A = -1
-tan A = 1
tan(pi - A) = 1 or tan(2pi - A) = 1
A = pi - tan^-1(1) or A = 2pi - tan^-1(1)
=3pi/4 or 7pi/4

but -pi < A <= pi

therefore A = -pi/4 or 3pi/4

but (2,-2), lying in quadrant 4,

hence A = -pi/4

therefore z = 2root2 cis -pi/4

 

[richz]

those are perfectly correct answers, y r u so worried for?

 

[sasquatch]

because its weird how im doing it...and the sheet doesnt explain it or anything

could someone please give me an full example of how they would go about solving a question such as above?

 

[richz]

errrr...... ur way. No difference,

i would just draw it, find the mod and arg. Then look at it and if it is under the x axis add a negative infront of the arg

 

[sasquatch]

well my thing works yeah.....but look at this question....

z = 3 + 4i

|z| = root(9+16)
=5

tan A = 4/3

tan A = 4/3 or tan(180+A) = 4/3
A = tan^-1(4/3) or A = tan^-1(4/3) - 180
= 53*8' or = -306*52'

but -180 < A <= 180

A = -126*52' or -486*52'

but look it screwed!!!!!!!!

the answer is 53*8 but my thing didnt work just now

your method sounds good but what do you mean add a sign infront of the argument? can you give me an example...

can you do this question z = -3 + root3 i then

OR check this out..

if y is negative then take away 180 from the angle....but if its positive just leave as is

 

[richz]

huh?? wat are u doing, the vector is in the first quad y do u have to get the other angles??? just use the angle u obtained.

(yes for those qs put it in degrees, but whenever u can put them in rads.)

Well when working our the arg, i ignore all negatives just work out the angle then if it is under the xaxis, obviously it would be a -ve. and if its in quads other than 1 or 4, subtract that angle from pi.

ok

|z|= sq root (9+3)
=root(12)
=2root3

then argz=tan^-1(root3/3)
argz=pi/6

obviously the vector is in quad 2 so we use (pi-@) = 5pi/6

so z=2root3cis(5pi/6)

(hopefully no mathematical errors)

 

[sasquatch]

so your saying


work out the angle ignoring all negatives

if the angle is in quadrant 1 leave as is
if the angle is in quadrant 2 the actual angle is 180 - angle (or pi - angle)
if the angle is in quadrant 3 do the same as above THEN wack a negative sign infront of it
if the angle is in quadrant 4 wack a negative sign infront of it?

 

[Riviet]

so your saying


work out the angle ignoring all negatives

if the angle is in quadrant 1 leave as is
if the angle is in quadrant 2 the actual angle is 180 - angle (or pi - angle)
if the angle is in quadrant 3 do the same as above THEN wack a negative sign infront of it
if the angle is in quadrant 4 wack a negative sign infront of it?

I agree with you on quadrants 1 & 2 & 4, but an alternative way of working the angle in the 3rd quadrant is by subtracting 360 degrees/2pi from the angle.

 

[sasquatch]

i just checked that with a question

z = -3 -2i

Arg(z) = tan^-1(2/3)
=33*41'

but if you subract 360 then you get -326*19' which is obviously out of the range -180 < A <= 180

but with what i said...

-(190 - 33*41')
= -146*19'

is in the range AND the correct answer.. just yeah..

 

[richz]

so your saying


work out the angle ignoring all negatives

if the angle is in quadrant 1 leave as is
if the angle is in quadrant 2 the actual angle is 180 - angle (or pi - angle)
if the angle is in quadrant 3 do the same as above THEN wack a negative sign infront of it
if the angle is in quadrant 4 wack a negative sign infront of it?

yes,

sas, buddy u need to stressless, its just the hsc, its not the end of the world.

 

[sikeveo]

All Stations to Cabra

We were told its "All sluts take cash"

 

[richz]

lmao, maybe too rude for my teachers to teach us that, lol.

 

[sasquatch]

oh well thanks for all your help xrtzx..