Roots of Complex Numbers (2 Viewers)

[Grey Council]

I'm a bit confused on how to do the roots of complex numbers. I mean, i can do the basic ones, like
(sqrt.3 +i)^6

But i have trouble with with
(-1+i)^6
and
(3-4i)^3

With the second one, i get like cos(-225) or whatever, but i don't know how that works. Do i measure it clockwise, so that it lies in the second quadrant? And the second one, it doesnt have to be exact ratios, right? So its like 54 degrees, but once again, as it lies in the fourth quadrant, what happens?

Sorry, i'm really confused.

 

[wogboy]

When finding powers of complex numbers follow these steps (De Moivre's Theorem):

e.g. (a + ib)^n

1. Convert to polar form (i.e. r*cis@ form)
2. Raise r to the power of the power n, and multiply @ by n, so you end up with r^n * cis(n@).
3. Convert back to rectangular form (i.e. x + iy form) if required.

----------------------------------------

e.g. for (3-4i)^3,

3 - 4i = 5 cis(-arctan(4/3))
(3-4i)^3 = 5^3 * cis(-3*arctan(4/3))
= 125 *[cos(3*arctan(4/3)) - i*sin(3*arctan(4/3))]
= 125 * [-0.936 - i*0.352]
= - 117 - 44*i

I'll let you try the other one.

 

[Grey Council]

hmph

Hey

I know the steps, its just that i got confused as to what differnece it makes if the original equation is in the first quadrant, second quadrant, third quadrant etc. No difference?

e.g. for (3-4i)^3,

3 - 4i = 5 cis(-arctan(4/3))
(3-4i)^3 = 5^3 * cis(-3*arctan(4/3))
= 125 *[cos(3*arctan(4/3)) - i*sin(3*arctan(4/3))]
= 125 * [-0.936 - i*0.352]
= - 117 - 44*i

(-1+i)^6
-1+i = sqrt 2 CIS (-pi/4) <---- Is that right? is it - pi on 4? or what? Cuase -pi/4 is positive/negative for sin/cos in different quadrants.
(-1+i)^6 = 2^3 CIS (-6 * pi/4)
= 8 (cos (-3pi/2) - i*sin(3pi/2)) <---- is it sin -3pi/2? cuase angle is in second quadrant, and therefore positive for sin?

I hope you see what i mean. I am befuddled as to how it alters when you get:
r^n *( cos(n*delta) -i * sin (n*delta)) when n*delta, for example, is positive for sin but negative for cos, ie lies in second quadrant, or for example n*delta is positive for cos but negative for sin, ie lies in the fourth quadrant.

If you don't know what i mean, i can try and clarify it further.

 

[wogboy]

I know the steps, its just that i got confused as to what differnece it makes if the original equation is in the first quadrant, second quadrant, third quadrant etc. No difference?

None at all
(r*cis@)^n = r^n * cis(n@) is always true.

-1+i = sqrt 2 CIS (-pi/4) <---- Is that right?

No it's actually sqrt(2)*cis(3*pi/4). Here's some important rules worth remembering for a complex number r*cis@:

- In the top right quadrant, 0 < @ < pi/2
- In the top left quadrant, pi/2 < @ < pi
- In the bottom right quadrant, -pi/2 < @ < 0
- In the bottom left quadrant, -pi < @ < -pi/2
- @ increases as you go anti clockwise.
- @ is always expressed between -pi and +pi (i.e. it's improper to say cis(3*pi), say cis(pi) instead). Obviously cis@ = cis(@ + 2*pi) for all @.

 

[Giant Lobster]

one question; is argz =pi<@<=pi or is that Argz? (i.e. which is principle)

I knew it as Argz being principle; argz being anything, but my teacher goes otherwise.. hmmm

 

[Grey Council]

oh!

No it's actually sqrt(2)*cis(3*pi/4).

OOO! Thats right, you measure from the 0 line, right? hehee, i get it now, i think. Thanks heaps.

Quite simple, innit? Thing is, i'm teaching myself, and i only have photocopies of the fitzpatrick book (with no answers) and the worked examples on the powers of complex numbers is missing in the photocopy. lol, just my bad luck, mm?

I have a photocopy of the Cambridge section of Complex Numbers as well, but the cambridge doesnt have worked examples on ANYTHING, but it does have the answers. So basically, im royally pucked, as the cambridge one is uber hard'ish. I'll finish off the fitzpatrick one and then start on the cambridge one again. Maybe i'll understand it better.

ANYWAY, thanks for the help, I really appreciate it. Any other pointers? Should i be doing fitzpatrick/cambridge? Anything else at all?

 

[Grey Council]

Im sorry, Lobster, I can't make head or tails of your question. Could you please rephrase it?

 

[mercury]

Arg z is principle argument, from - pi to pi
arg z is the 2kpi one, ie. anticlockwise from 0, going round and round and round.

arg z = Argz + 2kpi

 

[Giant Lobster]

Yes, that confirmed what I thought was the case. Thanks

 

[Grey Council]

can someone please confirm for me that 31(f) is the last exercise in the fitzpatrick book for complex numbers? ta

 

[Giant Lobster]

Yes thats correct.

Regarding that book however, it has jack all on complex number vector operations. Its very important u get familiar with vectors n stuff, which is largely neglected in that book.

 

[Grey Council]

o!

Which book is it in? Will i find it in the excel book (the old one)?

 

[Grey Council]

BTW, here are some more questions i couldn't do:

1. z is a complex number where the modulus of z = 1 and Argument of z = A. Find, in terms of A the value of Arg(z+a). Justify your answer.

2. a) Solve z^5 over the complex field (can do)
b) If w is the complex root of z^5 = -1 with the smallest positive argument, show that the other complex roots equal -w^2, w^3 and -w^4 (can do as well)
c) Using w as in part b, simplify (1 - w + w^2 - w^3)^8 (Can't do this one)

3. Describe and sketch on an argand diagram the locus of z given that Re(z - 1/z) = 0 where z is not equal to zero.
For this i get x^2 - y^2 = 1. This isnt a circle, is it? is it called a hyperbola? And how would i describe this wierd shape?

Thanks for your time.

 

[ND]

1. Mind-blank.

2. c) z^5=-1
z^5+1=0
roots of this eqn are -1,w,-w^2,w^3,-w^4
sum of roots:
-1+w-w^2+w^3-w^4=0
1-w+w^2-w^3=-w^4
(1-w+w^2-w^3)^8=(-w^4)^8
=w^32=w^30.w^2=(w^5)^6.w^2=w^2

3. Yep this is a hyperbola (rectangular hyperbola actually).

 

[wogboy]

1. z is a complex number where the modulus of z = 1 and Argument of z = A. Find, in terms of A the value of Arg(z+a). Justify your answer.

Where did the "a" in Arg(z+a) come from?

3. Describe and sketch on an argand diagram the locus of z given that Re(z - 1/z) = 0 where z is not equal to zero. For this i get x^2 - y^2 = 1. This isnt a circle, is it? is it called a hyperbola? And how would i describe this wierd shape?

I got a different answer actually.

z = x + iy
Re(z - 1/z) = 0
Re(z) - Re(1/z) = 0
Re(z) = Re(1/z)
Re(x + iy) = Re{(x - iy)/(x^2 + y^2)}
x = x/(x^2 + y^2)

NB: Never divide both sides of an equation by x unless you are absolutely certain that x =/= 0.

x * (x^2 + y^2) = x
x * (x^2 + y^2 - 1) = 0

hence x^2 + y^2 - 1 = 0 (i.e. x^2 + y^2 = 1), OR x=0.

So I believe the locus should be the circle x^2 + y^2 = 0 AND the line x=0 (i.e. the y-axis)

Don't forget to exclude the origin (i.e. the point (0,0)) from your sketch since you're explicitly told z =/= 0.

So you can express the locus like this:
(x^2 + y^2 = 1) UNION (x = 0 INTERSECTION y=/=0)

 

[Grey Council]

Ooops, sorry about the typo. The questions is:

1. z is a complex number where the modulus of z = 1 and Argument of z = A. Find, in terms of A the value of Arg(z+1). Justify your answer.

Thanks for the prompt replies. Really appreciate it.

BTW, howd you get so good at this Wogboy? You finished year 12 this year?

 

[wogboy]

1. z is a complex number where the modulus of z = 1 and Argument of z = A. Find, in terms of A the value of Arg(z+1)

For this question all you need to use is a bit of geometry. First plot z onto a Argand diagram (call this point A), then plot z+1 (call this point B). Call the origin O.

since you're given |z| = 1.
OA = 1
since (z+1) is 1 unit right of z,
AB = 1
so AB = OA
therefore triangle AOB is isosceles,
angle AOB = angle ABO

since AB is parallel to the x-axis,
angle OAB = pi - A (co-interior angles)

so angle AOB = A/2 (angle sum of a triangle is pi radians)
but angle AOB = arg(z) - arg(z+1)
so arg(z) - arg(z+1) = A/2
A - arg(z+1) = A/2

arg(z+1) = A/2

BTW, howd you get so good at this Wogboy? You finished year 12 this year?

Did the HSC last year (in 2002), currently studying elec eng at UNSW.

 

[Grey Council]

meh, and i thought i was getting the hang of complex numbers. :|

I've finished the fitzpatrick exercises, but now that i'm starting to do the exams, i'm finding that the exam questions are much harder. In your opinion, would i be better off just doing as many exams as possible (i've already printed out around 16 of the papers from Buchanan's site, will most likely print around 80, following KeypadSDM opinion) or I should do the Arnold book first? Thing is, before doing the fitzpatrick book, i tried the arnold one, but it seemed way too hard. Now I'm not sure if i should do the arnold one, cause i can just do previous trial papers and maybe the excel book. Whats your opinion?

Once again, thanks for your time and your patience in answering my questions. ^_^

 

[Grey Council]

here goes

Coroneos is best. You can get Coroneos' books through bookshops, or P.O.Box 25, Rose Bay, NSW, 2029

You mean coroneos is best for ALL the topics? ie, best for complex numbers, graphs, etc. Regardless, i'm ordering it in around 15 minutes. Hehe, Buchanan knows his shit, right? (No disrespect intended)

For complex numbers, try http://www.geocities.com/coroneosonline , exercises 01-13, 46-49 - but the books are better.

I'll print those exercises as soon as i finish this post.

Do past hsc papers 1990-2003 all questions, and the trials and specimen papers on my website do q8s only. For hsc exams prior to 1990 though do only questions in the syllabus.

What the hell! You mean i printed all those papers for nothing? DAMN IT! lol, so I should do ONLY questions 8 from those previous trial papers from other schools?

I reckon if you do it this way you can get 100 and beat keypad (albeit by 1 mark).

uhuh. Don't go too far.

Didn't your teacher give you stuff to do in the holidays? If not, you will find plenty on my website. There are currently 188 four unit papers on my website, amongst other things.

Umm, you know what an incompetent teacher is? Well, my teacher can't even be called a teacher. Qualifications: General maths in high school, diploma of teaching. Thats it. Has been stuffing up maths students for the past 20 years, although this year, and i quote: "I'll give 4u maths a try". In short, she doesn't know 2u maths, let alone 4u maths.

But yeah, I discovered your site. I think i've been hogging all the bandwidth on your site for the past few days. Download as much as i can, till it says bandwidth exceeded. lol

What school do you go to?

Sydney Boys High School. Its decent.

 

[Grey Council]

Originally posted by drbuchanan
GuardiaN,

You sound like a very good student and your teacher sounds pathetic.

You didn't waste time printing the papers. Save them for later when you revise for the trial.

Keypad worked hard to come first. Let me quote an ancient mathematician, Euclid, "there is no royal road to geometry", and a modern mathematician, Andrew Wiles, "there is simply no substitute for hard work". So for thousands of years, hard work is what gets results in maths.

hehee, there is hard work, and then there is a guy called Ivan Guo, Bronze medal Olympian, who is:
a) Chinese (<--- lol, its a joke)
b) Absolutely crazily psycho at maths.

He doesnt study though, he couldnt do a time payments question in our 2u assessments. But then he topped the 3u assessment. Hmm, if i work hard i reckon i can get top 10 in my grade. Cause there is the teacher thing as well, and the fact that my mind is more inclined to the english subject side. But yeah, i'm trying to be a good student. Its only one year, so I might as well give it the best shot I can.

As for the trial papers, after doing the coroneos exercises, what else should i do for complex numbers? nothing? I've started graphs, and it doesnt seem very hard (at the moment), so should i just continue on with the graphs after the coroneos exercises?