math1001: differential calculus (3 Viewers)

[Winston]

Originally posted by Wyvern
I already told you, you convert -1+i into polar form. Then you cube root it (cube rooting a poal complex number is simple and straight from the notes). That will be your first solution.

To get the other solution you add (2*PI)/3 to the angle and then again to get the third solution. Its straight from the notes man.

i thought you get an angle of -Pi/4, given that a = -1 and b = 1, tan = b/a = -1/1 = -1 = -45 = -Pi/4, and given the general rule

-Pi/4 +/- 2kPi, where k is consecutive, e.g. 1,2,3

 

[Collin]

Our first question is to sketch the set A, where A is the element of the complex number set, such that [mod(z-1)]=1.

Probably a typical question 2 for the regular classes.

 

[Wyvern]

Originally posted by Winston
i thought you get an angle of -Pi/4, given that a = -1 and b = 1, tan = b/a = -1/1 = -1 = -45 = -Pi/4,

Well thats part of converting it into polar form

and given the general rule

-Pi/4 +/- 2kPi, where k is consecutive, e.g. 1,2,3

I am not sure if your method works but

Since the n nth roots are evenly spaced around a circle, they are separated by an angle of (2* PI)/n

 

[Wyvern]

Originally posted by JKDDragon
Our first question is to sketch the set A, where A is the element of the complex number set, such that [mod(z-1)]=1.

Probably a typical question 2 for the regular classes.

Yeah in second year

nah our 3a is sort of like that and it is probably the hardest question in the assignment

 

[Collin]

Really? They give you examples of these kinds of sketch 'mod(z-a)=k' questions at around page 20 in the book. Should be really, really easy, even for non-advanced students.

 

[Carlito]

man, can someone explain to me how to do 3(a) parts ii and iii

 

[Wyvern]

Originally posted by Carlito
man, can someone explain to me how to do 3(a) parts ii and iii

for part ii) Re z >= -1 you can figure out by looking at page 16 on your notes at example v). It would be virtually the same thing only its the whole area greater then x=-1 as it is asking for Re z instead of Imz.
the upside down U thing means sketch the area included in both Rez >= -1 and arg z between PI/4 and PI2.

Now arg z = PI/2 is the positive y axis and arg z = PI/4 is the line y=x so it is basically the area between those 2 lines as they are both included in Rez >=-1

 

[lm1122]

I'm sooooo confused.... maybe i shouldn't have slept the last few lectures... well, i mean 8am is just rediculous. Really. But anyway, i dont know if i'm right or not, but for Q1(b)iii i got 2root8cis(13pi/12)... if it is right (and im really scared its not) how do we get 13pi/12 to fit in the given domain? (- pi < O < pi).... Ahhhhhh i hate maths!

 

[Wyvern]

Really. But anyway, i dont know if i'm right or not, but for Q1(b)iii i got 2root8cis(13pi/12)... if it is right (and im really scared its not) how do we get 13pi/12 to fit in the given domain -pi<0<pi???????????????????? Ahhhhh i hate maths [/B]

well go the other way.........13PI/12 is equal to -11PI/12

 

[lm1122]

omg! that completely makes sense! thanks!

 

[Winston]

Originally posted by lm1122
I'm sooooo confused.... maybe i shouldn't have slept the last few lectures... well, i mean 8am is just rediculous. Really. But anyway, i dont know if i'm right or not, but for Q1(b)iii i got 2root8cis(13pi/12)... if it is right (and im really scared its not) how do we get 13pi/12 to fit in the given domain? (- pi < O < pi).... Ahhhhhh i hate maths!

I'm not sure how you're getting that answer for 1b (iii) because it's totally different for me, how are you doing the working out for that ques? Polar Multiplication?

 

[Carlito]

Originally posted by Winston
I'm not sure how you're getting that answer for 1b (iii) because it's totally different for me, how are you doing the working out for that ques? Polar Multiplication?

yeah use the results from parts i and ii and perform polar multiplication on them

 

[Carlito]

for 3c did you get (x^2 - 2x + 2) as the answer from the long division?

if so, what do we do after that to get the roots, because i cant factorise that... or do i have to use the dodgy quadratic formula?

 

[Winston]

Originally posted by Carlito
for 3c did you get (x^2 - 2x + 2) as the answer from the long division?

if so, what do we do after that to get the roots, because i cant factorise that... or do i have to use the dodgy quadratic formula?

yea use the quadratic, and you should have two complex roots.

 

[JUB JUB]

yep that's right.

and yes you have to use the quadratic formula....but it works out nicely

 

[Wyvern]

Anyone got any hints about 3 a) iii) ?

 

[Winston]

Originally posted by Wyvern
Anyone got any hints about 3 a) iii) ?

ok... well look at Rez + Im z < 0 as

x + y < 0
y<-x

mmmm

 

[JUB JUB]

just graph them seperatly and then join them.... =)

it kinda resembles a downhill road wiith a road hump lol

 

[Winston]

Originally posted by JUB JUB
just graph them seperatly and then join them.... =)

it kinda resembles a downhill road wiith a road hump lol

yay i got it right

 

[Carlito]

Im completley lost with that question, its the only one i have left to do lol