HSC Complex Number Questions (1 Viewer)

[kayven]

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]

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]

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]

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!

 

[SpiralFlex]

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!

 

[Alkanes]

LOL T_T gimme a break mate

 

[bleakarcher]

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]

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]

LOL T_T gimme a break mate

Have a Kit-Kat

 

[SpiralFlex]

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.

 

[kayven]

i tried to use AAEldar's first method but i only got 1 solution: 2(cos(0)+isin(0))

 

[SpiralFlex]

i tried to use AAEldar's first method but i only got 1 solution: 2(cos(0)+isin(0))

(Keep this aside.)








or

or

 

[kayven]

ohhh i see now. Thank you so much.

 

[_zhangstaa]

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]

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