Question on Complex Numbers in Mathematics

[dpbird90]

Hi all, it's that time again, I'm doing maths and it isn't going well, so I need the help of the collective PH mind. This term's topic is Complex Numbers, and I am working on some questions from a sheet. I'm stuck on a couple of questions. The first being:

"solve the quadratic equation z^2+2iz+(116-44i)=0. Test one of your solutions by substituting it for z. Solutions should be complex numbers, so no calculators". Having gone through my lecture notes with a fine tooth comb, I still have no idea where to begin. Any thoughts?

The other question I'm stuck on is:

"Choose a very large positive integer n (1,000,000 is large enough). Make an estimate (no calculators!) for the principal argument of the complex number 1+[(pi*i)/n]. A good drawing is required (argument and diagram). The estimate should be so sharp that you can still use it to get a good estimate for the complex number {1+[(pi*i)/n]}^n.

The second estimate should follow automatically from the first. For comparison, offer an interesting estimate (with or without calculator) for the real number (1+[pi/n])^n n.b. don't just write 23.1406926... because this is not interesting (apparently)"

Sorry its a bit long to read, but if anyone can help I'll greatly appreciate it, virtual pints all round if you can help me solve it

[physprof]

its late and i'm an experimentalist... but the first one roots of a quad eqn.. maybe start from -b plus or minus sq root b squared - 4ac all over 2a...?

bks to maths actually....

[stuthemong]

for the first one I think you'd have to substitute z= a+bi

then put that into the equation (i.e. multiply it out)

then compare real + imaginary parts to solve for a and b

[Shay HTFC]

2 to the power of 2 is equal to 2 x 2 which is also equal to 2 + 2

I hope that helps

[dudleybloke]

not had to do anything like this for 18 years now.

X = fuktifino

[ScottishSamurai]

7


the confusing stuff always ends up as 7...

[Blue Meanie]

Wibble

[Tonto]

Answer to the first one is 42.
As for the second sum, my brain exploded just reading it.

[dilbert]

I have to go to a job centre interview (first thing) tomorrow, so I'm not qualified to answer your question and I have to sleep!

The answer, if it helps;

z = -2-12 i

and

z = 2+10 i

According to:

http://www.wolframalpha.com/input/?i=z^2%2B2iz%2B%...

It probably won't help, and anyway, the point of the question was to destroy the ambition of the thick and make the smart lazy, so I'm not entirely clear why you're asking here!!!

Edited by dilbert on Tuesday 16th February 01:56

[Lost_BMW]

dpbird90 said:

Hi all, it's that time again, I'm doing maths and it isn't going well, so I need the help of the collective PH mind. This term's topic is Complex Numbers, and I am working on some questions from a sheet. I'm stuck on a couple of questions. The first being:

"solve the quadratic equation z^2+2iz+(116-44i)=0. Test one of your solutions by substituting it for z. Solutions should be complex numbers, so no calculators". Having gone through my lecture notes with a fine tooth comb, I still have no idea where to begin. Any thoughts?

Sorry its a bit long to read, but if anyone can help I'll greatly appreciate it, virtual pints all round if you can help me solve it

Too late to look through/ for the 2nd right now but for the 1st maybe this will help . .

z 2 + 2 i z + ( 116 - 44i ) = 0

re-ordered to:

2 i z + z 2 + 116 = 44i

so as just one worked through test z = 16 and i = 31 works out . .

2 ( 31 x 16 ) + 162 + 116 = 44 x 31

=> 2 ( 496 ) = 992

992 + 256 = 1248

1248 + 116 = 1364

1364 / 44 = 31

sorry, this is very clumsily re-structured / written but I should be asleep (woke up in chair at 12.15, came to bed and made fatal mistake of just having a quick look in the Lounge!) Do check me ‘cos I’m braindead now.

I tried a couple of guesses based on even values for the substituted z and thought 10 would work + another that looked to work but did cheat(!) by using a calculator for ploughing through the substitutions and it didn’t come out right?

EDITED to make the square (2 s) small text as I'd been working in word and the superscript format hasn't come over.

Edited by Lost_BMW on Tuesday 16th February 02:30

[Lost_BMW]

Doh, I am tired!

I nearly gave in on 10 but it felt like it should work so I couldn't resist a few more trials and substitutions, but the tries I did all seemed to give the wrong 'balance' between i and z (the i seemed too large and so the 44 x i larger than the result of the 2(i x 10) + 100 + 116.

I just about gave in and thought 'off to bed' then realised I'd been going the wrong way and a couple of tests later got z = 10 and i = 9 to work out

2 i z + z2 + 116 = 44i

2 ( 9 x 10 ) + 102 + 116 = 44 x 9

=> 2 ( 90) = 180

180 + 100 + 116 = 396

44 x 9 = 396

Edit x 2 = double doh!! Didn't read your post correctly and spotted (leapt on) the word substitute completely missing the point about complex numbers etc. Sorry - glad more aware folk have helped and removed the risk of my cocking it all up for you

Edited by Lost_BMW on Tuesday 16th February 17:04

[Murdoc]

I'm a bit rusty but i=sqrt(-1) so your not looking for the solution of i it's imaginary. You need to square everything so you can get the (sqrt(-1))^2=-1 out then solve it as a normal quadratic. I will have a better look when I get to work.

EDIT: Looking at it again you don't want to lose the imaginary part.

Edited by Murdoc on Tuesday 16th February 08:18

[dpbird90]

Murdoc said:

I'm a bit rusty but i=sqrt(-1) so your not looking for the solution of i it's imaginary. You need to square everything so you can get the (sqrt(-1))^2=-1 out then solve it as a normal quadratic. I will have a better look when I get to work.

Yeah that's right, i=sqrt(-1). I'm not sure about substituting z=a+bi, as then I have 2 more unknowns to find (z,a,b) and I'll be even more confused. I've got the lecture at 9, so I'll double check with the lecturer (especially on that second one!) but in the meantime any more help is greatly appreciated

[stuthemong]

do. The substitution and you'll solve it. You can take z out with subst then find a and b by comparing real and imag. Parts. Then z is a +ib

iPhone typos

[Murdoc]

The answer from dilbert seems correct.

re-arrange to z^2+2iz-1 +117-44i=0

Treat the first bit as a quadratic so you can get:

(z+i)^2 +117 -44i=0

Then you can plug it into the normal quadratic eqn.

(-b+-sqrt(b^2-4ac)/2a

[dpbird90]

Murdoc said:

The answer from dilbert seems correct.

re-arrange to z^2+2iz-1 +117-44i=0

Treat the first bit as a quadratic so you can get:

(z+i)^2 +117 -44i=0

Then you can plug it into the normal quadratic eqn.

(-b+-sqrt(b^2-4ac)/2a

Thanks, I'm down to putting it into the quadratic equation, just one quick question, which coefficients do I take as a,b,c?
Thanks to dilbert too

ETA if anyone's interested, had a conversation with the lecturer earlier, the second question involves drawing a circle with radius 1 on a pair of axes. He then suggested I take n as 30, as then for the purpose of the diagram, the y co-ordinate would be approx 1/10 (as a complex number a+bi can be plotted on a pair of axes at the point (a,b). Then this forms a triangle, and he suggested I take an estimate of the modulus to be 1 (it is actually the hypotenuse of the the triangle, and a wee bit more than 1 but for an estimation I was told to take it as 1). Then there will be an arc of the circle within the triangle, and the length of this is pi/n approx. For the second bit where the equation is raised to the power n, the modulus is raised to the power n (no difference as its 1 anyway) and the argument is multiplied by n.

Edited by dpbird90 on Tuesday 16th February 10:50

[dpbird90]

anyone still there?

[blinkythefish]

dpbird90 said:

Hi all, it's that time again, I'm doing maths and it isn't going well, so I need the help of the collective PH mind. This term's topic is Complex Numbers, and I am working on some questions from a sheet. I'm stuck on a couple of questions. The first being:

"solve the quadratic equation z^2+2iz+(116-44i)=0. Test one of your solutions by substituting it for z. Solutions should be complex numbers, so no calculators". Having gone through my lecture notes with a fine tooth comb, I still have no idea where to begin. Any thoughts?

The other question I'm stuck on is:

"Choose a very large positive integer n (1,000,000 is large enough). Make an estimate (no calculators!) for the principal argument of the complex number 1+[(pi*i)/n]. A good drawing is required (argument and diagram). The estimate should be so sharp that you can still use it to get a good estimate for the complex number {1+[(pi*i)/n]}^n.

The second estimate should follow automatically from the first. For comparison, offer an interesting estimate (with or without calculator) for the real number (1+[pi/n])^n n.b. don't just write 23.1406926... because this is not interesting (apparently)"

Sorry its a bit long to read, but if anyone can help I'll greatly appreciate it, virtual pints all round if you can help me solve it

Dilberts answer is correct. An alternatve method:

(z+r1)*(z+r2)= z^2 + 2iz + (116 -44i), where r1 = a+bi and r2 = c+di

multiply out gives

z^2 + (a+c + (b+d)i)z + ((ac-bd)+(ad+bc)i)

Substituting back in gives four equations:

a+c = 0 (because there is no real componen times z.
b+d = 2 (because there is a 2i imaginary component.
ac-bd = 116
ad+bc = -44

which can be solved as a system of linear equations. a= -2, c=2 b=-12, d=10

The second part:

the principal arguement(theta) is the angle between the real axis and the vector described by the complex number:

for large +ve n, it forms a triagle where:

Opposite= pi/n, adjacent =1.

tan(theta)= O/A = (pi/n)/1 = pi/n. From memory, I seem to recal that tan(x) tends towards x as x tends to 0. i.e for a big n, theta tends to pi/n

edited to add reference:
http://en.wikipedia.org/wiki/Small-angle_approxima...


Also for n large enough, the hypotenuse is approx equal to Opposite, i.e 1.

As to raising it to n.
(1+i) has an angle(principal arguement) of pi/4
(1+i)^2 = 2i has an angle pi/2
(1+i)^3 = 2i-2 which is an angle of 3pi/4
....

and so on, raising to the power of n, represents a clockwise rotation of by the principal arguement. therefore (1+(pi/n)i)^n will have an angle of pi, which means the result is negative and real. I think the length of the vector can be left as 1, since the modulus of 1+(pi/n)i can be approxomated by 1, but in reality it describes a spiral(look at the 1+i example, the modulus goes sqrt(2), 2, 3sqrt(2) with each power)......

Edited by blinkythefish on Tuesday 16th February 14:12

Edited by blinkythefish on Tuesday 16th February 14:14

Edited by blinkythefish on Tuesday 16th February 14:16

[dpbird90]

blinkythefish said:

dpbird90 said:

Hi all, it's that time again, I'm doing maths and it isn't going well, so I need the help of the collective PH mind. This term's topic is Complex Numbers, and I am working on some questions from a sheet. I'm stuck on a couple of questions. The first being:

"solve the quadratic equation z^2+2iz+(116-44i)=0. Test one of your solutions by substituting it for z. Solutions should be complex numbers, so no calculators". Having gone through my lecture notes with a fine tooth comb, I still have no idea where to begin. Any thoughts?

The other question I'm stuck on is:

"Choose a very large positive integer n (1,000,000 is large enough). Make an estimate (no calculators!) for the principal argument of the complex number 1+[(pi*i)/n]. A good drawing is required (argument and diagram). The estimate should be so sharp that you can still use it to get a good estimate for the complex number {1+[(pi*i)/n]}^n.

The second estimate should follow automatically from the first. For comparison, offer an interesting estimate (with or without calculator) for the real number (1+[pi/n])^n n.b. don't just write 23.1406926... because this is not interesting (apparently)"

Sorry its a bit long to read, but if anyone can help I'll greatly appreciate it, virtual pints all round if you can help me solve it

Dilberts answer is correct. An alternatve method:

(z+r1)*(z+r2)= z^2 + 2iz + (116 -44i), where r1 = a+bi and r2 = c+di

multiply out gives

z^2 + (a+c + (b+d)i)z + ((ac-bd)+(ad+bc)i)

Substituting back in gives four equations:

a+c = 0 (because there is no real componen times z.
b+d = 2 (because there is a 2i imaginary component.
ac-bd = 116
ad+bc = -44

which can be solved as a system of linear equations. a= -2, c=2 b=-12, d=10

The second part:

the principal arguement(theta) is the angle between the real axis and the vector described by the complex number:

for large +ve n, it forms a triagle where:

Opposite= pi/n, adjacent =1.

tan(theta)= O/A = (pi/n)/1 = pi/n. From memory, I seem to recal that tan(x) tends towards x as x tends to 0. i.e for a big n, theta tends to pi/n

edited to add reference:
http://en.wikipedia.org/wiki/Small-angle_approxima...


Also for n large enough, the hypotenuse is approx equal to Opposite, i.e 1.

As to raising it to n.
(1+i) has an angle(principal arguement) of pi/4
(1+i)^2 = 2i has an angle pi/2
(1+i)^3 = 2i-2 which is an angle of 3pi/4
....

and so on, raising to the power of n, represents a clockwise rotation of by the principal arguement. therefore (1+(pi/n)i)^n will have an angle of pi, which means the result is negative and real. I think the length of the vector can be left as 1, since the modulus of 1+(pi/n)i can be approxomated by 1, but in reality it describes a spiral(look at the 1+i example, the modulus goes sqrt(2), 2, 3sqrt(2) with each power)......

Edited by blinkythefish on Tuesday 16th February 14:12

Edited by blinkythefish on Tuesday 16th February 14:14

Edited by blinkythefish on Tuesday 16th February 14:16

Thanks very much assignment complete and handed in. Which means I now have a free afternoon. No more lectures, assignments etc, day off from work. Time to chill with a bit of rfactor methinks