HSC Complex Number Questions (1 Viewer)

kayven · Oct 30, 2011

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 · Oct 30, 2011

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 · Oct 30, 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 $z^3-8=0$ and then factorise so you have $(z-2)(z^2+2z+4)$ and use the quadratic formula and turn those answers into mod-arg form.

SpiralFlex · Oct 30, 2011

Let $r(\cos \theta + i\sin \theta)=\sqrt[3]{}8$

$r^3(\cos \theta+i\sin \theta)^3=8$

By taking modulus of both sides,

$r^3=8$

$r=2$ (Remember this important result.)

$(\cos \theta+i\sin \theta)^3=1+0i$

By De Moivre's Theorem,

$(\cos 3\theta+i\sin 3\theta)=1+0i$

By equating real coefficients,

$\cos 3\theta = 1$

$\cos 3\theta = \cos 0$

$3 \theta = 0 + 2k\pi$ Where $k$ is any integer.

$\theta=\frac{2k\pi}{3}$

Now, $k=0,1,2$

$z1=2(\cos 0+i\sin 0)=2+0i$

$z2=2(\cos \frac{2}{3} \pi + i \sin \frac{2}{3}\pi)=-1+\sqrt{3}i$

$z3=2(\cos \frac{4}{3}\pi + i \sin \frac{4}{3}\pi)=2(\cos \frac{-2}{3} \pi + i \sin \frac{-2}{3}\pi)=-1-\sqrt{3}i$

Note: I gave you the Cartesian forms too!

SpiralFlex · Oct 30, 2011

Alkanes said:
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. $(-3^2)=-9$

Should be $(-3)^2=9$

AAEldar said:
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 $z^3-8=0$ and then factorise so you have $(z-2)(z^2+2z+4)$ and use the quadratic formula and turn those answers into mod-arg form.

You are meant to be studying for your HSC!

Alkanes · Oct 30, 2011

LOL T_T gimme a break mate

bleakarcher · Oct 30, 2011

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 · Oct 30, 2011

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 · Oct 30, 2011

Alkanes said:
LOL T_T gimme a break mate

Have a Kit-Kat

SpiralFlex · Oct 30, 2011

kayven said:
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.

bleakarcher said:
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 · Oct 30, 2011

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

SpiralFlex · Oct 30, 2011

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

$z^3-8=0$

$(z-2)(z^2+2z+4)$

$z=2=2(\cos 0+i\sin 0)$ (Keep this aside.)

$z= \frac{-2 \pm \sqrt{4-4(4)}}{2}$

$=\frac{-2 \pm 2\sqrt{3}i}{2}$

$=-1 \pm \sqrt{3}i$

$=2(-\frac{1}{2}+\frac{\sqrt{3}}{2}i)$ or $2(-\frac{1}{2}-\frac{\sqrt{3}}{2}i)$

$=2(\cos \frac{2}{3}\pi+i\sin \frac{2}{3}\pi)$ or $2(\cos \frac{-2}{3}\pi+i\sin \frac{-2}{3}\pi)$

$\therefore z=2(\cos 0+i\sin 0), 2(\cos \frac{2}{3}\pi+i\sin \frac{2}{3}\pi), 2(\cos \frac{-2}{3}\pi+i\sin \frac{-2}{3}\pi)$

kayven · Oct 30, 2011

ohhh i see now. Thank you so much.

_zhangstaa · Nov 4, 2011

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

$cos\theta = {\frac{\frac{4+4-(\sqrt{8+4\sqrt{3}})^{2}}}{8}}$

will it become $cos\theta = {\frac{\frac{4+4-({8+4\sqrt{3}})}}{8}}$ or $cos\theta = {\frac{\frac{4+4+({8+4\sqrt{3}})}}{8}}$

i know its a bit silly but im really confused cause the cosine rule says $cos\theta =\fraca\frac{a^{2}+b^{2}-c^{2}}{2ab}$ 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 · Nov 4, 2011

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

$cos\theta = {\frac{\frac{4+4-(\sqrt{8+4\sqrt{3}})^{2}}}{8}}$

will it become $cos\theta = {\frac{\frac{4+4-({8+4\sqrt{3}})}}{8}}$ or $cos\theta = {\frac{\frac{4+4+({8+4\sqrt{3}})}}{8}}$

i know its a bit silly but im really confused cause the cosine rule says $cos\theta =\fraca\frac{a^{2}+b^{2}-c^{2}}{2ab}$ 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

HSC Complex Number Questions (1 Viewer)

kayven

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