4 Unit Revising Marathon HSC '10 (1 Viewer)

[Dragonmaster262]

the equation:
|z-1-3i|+|z-9-3i|=10 corresponds to an ellipse in the argand diagram

(i) write down the complex number corresponding to the centre

(ii) write down the range of values of arg(z) for complex numbers z, corresponding to points on the ellipse

from hsc past paper =/,

Is the centre 5+3i? I think I miscalculated something.

 

[adomad]

center is correct

range of arg(z)

0<=arg(z)<=pi/2

 

[Gussy Booo]

Is the centre 5+3i? I think I miscalculated something.

Lets see your working out? [=

 

[adomad]

 

[Gussy Booo]

center is correct

range of arg(z)

0<=arg(z)<=pi/2

omg omg tell me how. lol. i wanted to see working out, so i could kno how to do it.

explain plzzzzzzzzzzzzzzz.

+ how u got the centre.

they're both right btw.

 

[adomad]

well the outer most point is 0+3i... the arg of this is pi/2
the smallest arg is the tanget to the bottom of the ellipse.. work that out .. centre is 5+3i so form a triangle with sides 5 (half of 10) and 4(the distance from the middle to the point 1+3i)
and then u see that the ellipse hits the point 5+0i... arg 5=0

 

[adomad]

new question!:

express

 

[adomad]

or if that is too easy...

 

[Gussy Booo]

new question!:

express

Looks to easy =/...

Um....If we express it in mod arg form we get --> 1cis(pi/3).
Now..according to De Moivre [1cis([i/3)]^8 = 1^8cis(8pi/3).
But this doesn't lie number the principle argument.
So if we play around..it becomes --> 1^8cis(5pi/3).
This furthermore gives:

cos5pi/3+isin5pi/3.
We then punch some numbers into the calculator...which i cbf doing.

 

[adomad]

Looks to easy =/...

Um....If we express it in mod arg form we get --> 1cis(pi/3).
Now..according to De Moivre [1cis([i/3)]^8 = 1^8cis(8pi/3).
But this doesn't lie number the principle argument.
So if we play around..it becomes --> 1^8cis(5pi/3).
This furthermore gives:

cos5pi/3+isin5pi/3.
We then punch some numbers into the calculator...which i cbf doing.

twaz a trick Q

 

[omgmynameislong]

sin+icos

not cos+isin

(sinpi/3 + icos pi/3)^8

=(cos pi/6 + isin pi/6)^8
=cis -4pi/6

cbf a+ib

 

[Gussy Booo]

sin+icos

not cos+isin

(sinpi/3 + icos pi/3)^8

=(cos pi/6 + isin pi/6)^8
=cis -4pi/6

cbf a+ib

HAHAHAH! OMGGGGGG! thats soo funny.
i remember looking at the screen and saying...wait..why are my sin and cos the other way around. Then i just switched em around.
Good Q. >.< I would so lose a mark there.

 

[omgmynameislong]

or if that is too easy...

in terms of tanx?

(sinx+icosx)^3

= (sinx)^3 + i3(sinx)^2cosx - 3sinx(cosx)^2 - i(cosx)^3

equating real and imaginary parts

sin3x= (sinx)^3 - 3sinx (cosx)^2
cos3x= 3(sinx)^2cosx - (cosx)^3

tan3x = [(sinx)^3 - 3sinx(cosx)^2]/[3(sinx)^2cosx - (cosx)^3]

divide by (cosx)^3

tan3x= [(tanx)^3 - 3tanx]/[3tanx -1]

 

[cutemouse]

Prove by mathematical induction that

 

[Gussy Booo]

Someone please do this.

 

[nonsenseTM]

why is that induction a 4u question?

 

[ninetypercent]

not sure if this is right...

EDIT: it is wrong. RTP part/ ooopss



For n =1
LHS = 1
RHS = (4 + 3)/6 x 1 = 7/6
LHS < RHS
therefore, true for n =1

Assume true for n = k


RTP true for n = k + 1



The statement is true for n = k+1, if true for n =k

therefore, by induction, the statement is true

EDIT: latex looked ugly, so i had to edit a number of times

 

[Trebla]

Hint: First prove that

 

[hscishard]

:O
What a way of saying that's wrong.

 

[Lukybear]

For sqrt(k+1) <= (8k + 5 )/6
Not really sure if its an applicable method:

Solving via quadratics:

Via graph and observation, k > 0.1948 for
hence sqrt(k+1) <= (8k + 5 )/6 for all n>=1

Using Induction
Assume True for n = k


Prove True for n= k+1


From assumption and proven above, i.e. via quadratics induction is true...