HSC Complex Number Questions (1 Viewer)

kayven

Member
Joined
Jul 8, 2011
Messages
69
Gender
Undisclosed
HSC
2012
My teacher has given me this year's complex number questions. Would you be able to do them so that i can check my solution?

a) Let w=2-3i and z=3+4i
i) Find (conjugate)w + z
ii) Find |w|
iii) Express w/z in the form a+ib, where a and b are real numbers.

c) Find, in modulus -argument form, all solutions of z^3=8

Thank you so much!
 

Alkanes

Active Member
Joined
May 20, 2010
Messages
1,417
Gender
Male
HSC
2012
i ) 2-3i + 3 +4i = 5 + i. Conjugate is 5-i

ii) Root 2^2 + (-3^2) = Root 13

iii) 2-3i/3+4i Rationalise it. Should equal -6/25 - 17i/25

c) Not bothered sorry i think that's roots of unity. Havn't cleaned that up yet haha
 

AAEldar

Premium Member
Joined
Apr 5, 2010
Messages
2,246
Gender
Male
HSC
2011
There's 2 ways to do part c.

First way - Convert 8 into mod-arg form and then take the cubed root of it, using De Moivre's theorem.
Second way - Subtract the 8 so you have and then factorise so you have and use the quadratic formula and turn those answers into mod-arg form.
 

SpiralFlex

Well-Known Member
Joined
Dec 18, 2010
Messages
6,960
Gender
Female
HSC
N/A
Let




By taking modulus of both sides,



(Remember this important result.)




By De Moivre's Theorem,




By equating real coefficients,





Where is any integer.




Now,








Note: I gave you the Cartesian forms too! :)
 
Last edited:

SpiralFlex

Well-Known Member
Joined
Dec 18, 2010
Messages
6,960
Gender
Female
HSC
N/A
i ) 2-3i + 3 +4i = 5 + i. Conjugate is 5-i

ii) Root 2^2 + (-3^2) = Root 13

iii) 2-3i/3+4i Rationalise it. Should equal -6/25 - 17i/25

c) Not bothered sorry i think that's roots of unity. Havn't cleaned that up yet haha
Also technical mistake.

ii) Root 2^2 + (-3^2) = Root 13

This result is not true.

Should be


There's 2 ways to do part c.

First way - Convert 8 into mod-arg form and then take the cubed root of it, using De Moivre's theorem.
Second way - Subtract the 8 so you have and then factorise so you have and use the quadratic formula and turn those answers into mod-arg form.
You are meant to be studying for your HSC! :)
 
Last edited:

bleakarcher

Active Member
Joined
Jul 8, 2011
Messages
1,509
Gender
Male
HSC
2013
spiralflex, in the first line of ur working for c) u say:
Let r=cos(x)+isin(x)=cube root[8]
then u say
r^3[cos(x)+sin(x)]=8
 

kayven

Member
Joined
Jul 8, 2011
Messages
69
Gender
Undisclosed
HSC
2012
whoa, Spiral Flex, your way seems so complicated. I just used what AAEldar said and factorsed z^3-8=0 then use quadratic formula for z. Apparently, i got 3 solutions, they were: 2(cos(-2pi/3)+isin(-2pi/3)),2(cos(2pi/3)+isin(2pi/3)) and 2(cos(0)+isin(0)).

Are these correct? Btw how do you type that working out? do you use a programme?
 

SpiralFlex

Well-Known Member
Joined
Dec 18, 2010
Messages
6,960
Gender
Female
HSC
N/A
whoa, Spiral Flex, your way seems so complicated. I just used what AAEldar said and factorsed z^3-8=0 then use quadratic formula for z. Apparently, i got 3 solutions, they were: 2(cos(-2pi/3)+isin(-2pi/3)),2(cos(2pi/3)+isin(2pi/3)) and 2(cos(0)+isin(0)).

Are these correct? Btw how do you type that working out? do you use a programme?
Latex. I like this method better since it involves trigonometry. I recommend you to understand AAEldar's first method too.

spiralflex, in the first line of ur working for c) u say:
Let r=cos(x)+isin(x)=cube root[8]
then u say
r^3[cos(x)+sin(x)]=8
Typo. :p
 
Last edited:

kayven

Member
Joined
Jul 8, 2011
Messages
69
Gender
Undisclosed
HSC
2012
i tried to use AAEldar's first method but i only got 1 solution: 2(cos(0)+isin(0))
 

_zhangstaa

Member
Joined
Sep 5, 2010
Messages
32
Gender
Undisclosed
HSC
N/A
shouldnt (a) part i be different?
ive got 2+3i+3+4i = 5+7i since its only the conjugate of w not the conjugate of (w+z).

has anyone done (b) part ii? i got a a negative value for cos theta, which scares me :S



will it become or

i know its a bit silly but im really confused cause the cosine rule says and i dunno if when you square the c term you would take its negative sign as well to make it positive or just leave it. but if i leave the negative sign then i have a negative result for cos.
 

_zhangstaa

Member
Joined
Sep 5, 2010
Messages
32
Gender
Undisclosed
HSC
N/A
shouldnt (a) part i be different?
ive got 2+3i+3+4i = 5+7i since its only the conjugate of w not the conjugate of (w+z).

has anyone done (b) part ii? i got a a negative value for cos theta, which scares me :S



will it become or

i know its a bit silly but im really confused cause the cosine rule says and i dunno if when you square the c term you would take its negative sign as well to make it positive or just leave it. but if i leave the negative sign then i have a negative result for cos.

ps: sorry for the dodgy fraction thingie, its my first time using it and i have no idea how to change it :S
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top