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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: hard on November 18, 2008, 08:21:00 pm

Title: hards methods i mean specialist :D questions
Post by: hard on November 18, 2008, 08:21:00 pm: anwway year has started and here is the first question regarding complex numbers:

write the following in the form of x+yi, where x and y are real numbers.
a) i⁹+i¹⁰

Z = (i²)⁴ x i +(i²)5
= -1⁴ x i + (-1²)⁵
=-i + (-i)
=-i - i
=-2i Thus R_e=0 while I_m=-2i

hopefully i did that right ???
Title: Re: hards methods i mean specialist :D questions
Post by: Glockmeister on November 18, 2008, 08:23:54 pm: I've whack that in my CAS calc and I can tell you the answer is wrong.

$z = i^{9} + i^{10}$

$z = i^{8}.i + (i^{2})^{5}$

$z = (i^{2})^{4}.i + (-1)^{5}$

$z = (-1)^{4}.i - 1$

$z = i - 1$

If they ask "in the form of x+yi", you do not need to state Re(z) and Im(z).
Title: Re: hards methods i mean specialist :D questions
Post by: hard on November 18, 2008, 08:24:23 pm: Quote from: Glockmeister on November 18, 2008, 08:23:54 pm
I've whack that in my CAS calc and I can tell you the answer is wrong.
oh shit what did i do wrong? can someone explain
Title: Re: hards methods i mean specialist :D questions
Post by: Glockmeister on November 18, 2008, 08:28:38 pm: have a look at what I've done ahead.
Title: Re: hards methods i mean specialist :D questions
Post by: hard on November 18, 2008, 08:32:00 pm: OH SHIT i see my mistake thanks glockmeister +1 coming your way
Title: Re: hards methods i mean specialist :D questions
Post by: Damo17 on November 18, 2008, 08:32:38 pm: HARD: what text book are you using?
Title: Re: hards methods i mean specialist :D questions
Post by: hard on November 18, 2008, 08:34:21 pm: Maths quest although i'm using a photocopy of chapter 3 from my teacher so i haven't got the real book. what about you damo?
Title: Re: hards methods i mean specialist :D questions
Post by: Damo17 on November 18, 2008, 08:41:08 pm: Maths quest also, that was question 3a from 3A, i read it and thought it was so familiar.
Title: Re: hards methods i mean specialist :D questions
Post by: hard on November 18, 2008, 08:42:29 pm: Quote from: Damo17 on November 18, 2008, 08:41:08 pm
Maths quest also, that was question 3a from 3A, i read it and thought it was so familiar.

yer that one :P Have you started your specialist course yet damo?
Title: Re: hards methods i mean specialist :D questions
Post by: Damo17 on November 18, 2008, 08:44:12 pm: Yeah, we started 2 weeks of transition yesterday. Only half my GMA class had the guts to tackle specialist.
Title: Re: hards methods i mean specialist :D questions
Post by: dekoyl on November 18, 2008, 08:53:16 pm: I'm jealous of the schools that offer this transition thing. I think it's a good 'kick start' for the subjects. It's harder to pick up a book and sit down to study in the holidays than it is to just start at school and continue in the holidays :(
Title: Re: hards methods i mean specialist :D questions
Post by: hard on November 18, 2008, 09:41:16 pm: Quote from: Damo17 on November 18, 2008, 08:44:12 pm
Yeah, we started 2 weeks of transition yesterday. Only half my GMA class had the guts to tackle specialist.

half lol what the hell! Only 6 kids are doing it in our school and i didn't do general advanced maths so i'm starting fresh with spesh. Seems good so far.

also i got another question to check if it is right: To damo this is q1h of chapter 3A

Using the imaginary number i, write down expression for the following
$\sqrt{-36/25}\right)$

x² = -36/25
x² = -1 x 36/25
x² = i² x 36/25
x = (plusminus) $\sqrt {i^2 x 36/25}$
x = (plusminus) i x 6/5

$therfore x = (plusminus) 6/5i$

EDIT: change 35>25
Title: Re: hards methods i mean specialist :D questions
Post by: shinny on November 18, 2008, 09:45:13 pm: Your first step is incorrect. By stating $x^2=-\frac{36}{25}$ , you're adding another answer in which is why you have the $\pm$

Try this:
$\sqrt{-\frac{36}{25}}$
$=\sqrt{-1} \times \sqrt{(\frac{6}{5}) ^{2}}$
$=i \times \frac{6}{5}$
$=\frac{6}{5}i$
Title: Re: hards methods i mean specialist :D questions
Post by: hard on November 18, 2008, 09:47:38 pm: Quote from: shinjitsuzx on November 18, 2008, 09:45:13 pm
Your first step is incorrect. By stating $x^2=-\frac{36}{25}$ , you're adding another answer in which is why you have the $\pm$

Try this:
$\sqrt{-\frac{36}{25}}$
$=\sqrt{-1} \times \sqrt{(\frac{6}{5}) ^{2}}$
$=i \times \frac{6}{5}$
$=\frac{6}{5}i$

but were did you get $=\sqrt{-1}$
Title: Re: hards methods i mean specialist :D questions
Post by: hard on November 18, 2008, 09:48:46 pm: Quote from: hard on November 18, 2008, 09:47:38 pm
Quote from: shinjitsuzx on November 18, 2008, 09:45:13 pm
Your first step is incorrect. By stating $x^2=-\frac{36}{25}$ , you're adding another answer in which is why you have the $\pm$

Try this:
$\sqrt{-\frac{36}{25}}$
$=\sqrt{-1} \times \sqrt{(\frac{6}{5}) ^{2}}$
$=i \times \frac{6}{5}$
$=\frac{6}{5}i$

but were did you get $=\sqrt{-1}$

oh sorry, multiplying $=\sqrt{-1}$ with the other part would not change the answer, yerp thanx shinjitsuzx
Title: Re: hards methods i mean specialist :D questions
Post by: shinny on November 18, 2008, 09:49:07 pm: That's the definition of i.
$i=\sqrt{-1}$

EDIT: Oh, you meant as in that way. And as for your first step, just remember that any polynomial to the power of n has n number of solutions (fundamental theorem of algebra), so you should know that stating something like what you had would add invalid solutions in some way.
Title: Re: hards methods i mean specialist :D questions
Post by: Glockmeister on November 18, 2008, 09:50:36 pm: Quote from: hard on November 18, 2008, 09:41:16 pm
Quote from: Damo17 on November 18, 2008, 08:44:12 pm
Yeah, we started 2 weeks of transition yesterday. Only half my GMA class had the guts to tackle specialist.

half lol what the hell! Only 6 kids are doing it in our school and i didn't do general advanced maths so i'm starting fresh with spesh. Seems good so far.

also i got another question to check if it is right: To damo this is q1h of chapter 3A

Using the imaginary number i, write down expression for the following
$\sqrt{-36/25}\right)$

x² = -36/25
x² = -1 x 36/25
x² = i² x 36/25
x = (plusminus) $\sqrt {i^2 x 36/25}$
x = (plusminus) i x 6/5

$therfore x = (plusminus) 6/5i$

EDIT: change 35>25

Ok with this, yes and no. You've actually kinda cheated because you have used an equation by letting the expression equal something when the question asked for an expression.

EDIT: Obviously shin has dived in with the answer.
Title: Re: hards methods i mean specialist :D questions
Post by: hard on November 18, 2008, 09:52:21 pm: Quote from: Flaming_Arrow on November 18, 2008, 09:50:54 pm
Quote from: hard on November 18, 2008, 09:47:38 pm
Quote from: shinjitsuzx on November 18, 2008, 09:45:13 pm
Your first step is incorrect. By stating $x^2=-\frac{36}{25}$ , you're adding another answer in which is why you have the $\pm$

Try this:
$\sqrt{-\frac{36}{25}}$
$=\sqrt{-1} \times \sqrt{(\frac{6}{5}) ^{2}}$
$=i \times \frac{6}{5}$
$=\frac{6}{5}i$

but were did you get $=\sqrt{-1}$

$=\sqrt{-1} \cdot \sqrt{\frac{36}{25}} = \sqrt{-\frac{36}{25}}$

ya lol got that
Title: Re: hards methods i mean specialist :D questions
Post by: Glockmeister on November 18, 2008, 09:55:28 pm: oh btw hard it would have been funny if you got rid of the s in hard in the topic title.
Title: Re: hards methods i mean specialist :D questions
Post by: hard on November 18, 2008, 10:18:07 pm: Quote from: Glockmeister on November 18, 2008, 09:55:28 pm
oh btw hard it would have been funny if you got rid of the s in hard in the topic title.

:p haha i thought the same thing right before i clicked post
Title: Re: hards methods i mean specialist :D questions
Post by: danieltennis on November 18, 2008, 10:23:06 pm: Quote from: hard on November 18, 2008, 10:18:07 pm
Quote from: Glockmeister on November 18, 2008, 09:55:28 pm
oh btw hard it would have been funny if you got rid of the s in hard in the topic title.

:p haha i thought the same thing right before i clicked post

HARD, did u do GMA this year?
Title: Re: hards methods i mean specialist :D questions
Post by: shinny on November 18, 2008, 10:27:21 pm: Quote from: hard on November 18, 2008, 09:41:16 pm
...i didn't do general advanced maths so i'm starting fresh with spesh...

etcetc.
Title: Re: hards methods i mean specialist :D questions
Post by: Flaming_Arrow on November 18, 2008, 10:29:49 pm: i did it last year but don't remember anything
Title: Re: hards methods i mean specialist :D questions
Post by: hard on November 18, 2008, 10:53:10 pm: okay the question asks

evaluate:

R_e(-5+4i) and I_m(1-6i)
Title: Re: hards methods i mean specialist :D questions
Post by: gfb on November 18, 2008, 10:56:41 pm: Quote from: hard on November 18, 2008, 10:53:10 pm
okay the question asks

evaluate:

R_e(-5+4i) and I_m(1-6i)

Real = -5
Imaginary = -6

Not sure of what the Q asks :P
Title: Re: hards methods i mean specialist :D questions
Post by: hard on November 18, 2008, 11:03:35 pm: ahkai i was thinking that they were asking for more
Title: Re: hards methods i mean specialist :D questions
Post by: ell on November 18, 2008, 11:25:38 pm: Another tip you might find useful with complex numbers (you might already know this but I'll post it anyway):

When you have something like $i^{38}$ , an easy way to simplify it is to divide the power by 4, and whatever the remainder is becomes the new power (in this case, the remainder is 2 and so: $i^{38}=i^{2}=-1$ )
Title: Re: hards methods i mean specialist :D questions
Post by: hard on November 18, 2008, 11:55:16 pm: Quote from: ell on November 18, 2008, 11:25:38 pm
Another tip you might find useful with complex numbers (you might already know this but I'll post it anyway):

When you have something like $i^{38}$ , an easy way to simplify it is to divide the power by 4, and whatever the remainder is becomes the new power (in this case, the remainder is 2 and so: $i^{38}=i^{2}=-1$ )
thanks for that but i usually just put (i^2) and then put the remaining value outside the brackets needed for 2 to be multiplied by that number to equal in this case 38. This way i cancel the i^2 out and then have -1^16 in this case.but thanks again
Title: Re: hards methods i mean specialist :D questions
Post by: NE2000 on November 19, 2008, 02:50:09 pm: Ahh...so the baton passes on to 2009 spesh students. I've just started with the graphs in chapter one, pretty cool stuff so far...
Title: Re: hards methods i mean specialist :D questions
Post by: Mao on November 19, 2008, 03:07:02 pm: Quote from: shinjitsuzx on November 18, 2008, 09:45:13 pm
Your first step is incorrect. By stating $x^2=-\frac{36}{25}$ , you're adding another answer in which is why you have the $\pm$

Try this:
$\sqrt{-\frac{36}{25}}$
$=\sqrt{-1} \times \sqrt{(\frac{6}{5}) ^{2}}$
$=i \times \frac{6}{5}$
$=\frac{6}{5}i$

I would be very careful factoring that square root of -1

$1=\sqrt{1} = \sqrt{-1\times -1} = \sqrt{-1}\times \sqrt{-1} = -1$

as you can see, it's not exactly the 'right' thing to do in surds

what 'should' be done is $\sqrt{-\frac{36}{25}} = \sqrt{i^2 \times \frac{36}{25}} = \frac{6}{5}i$
Title: Re: hards methods i mean specialist :D questions
Post by: hard on November 19, 2008, 03:11:43 pm: Quote from: Mao on November 19, 2008, 03:07:02 pm
Quote from: shinjitsuzx on November 18, 2008, 09:45:13 pm
Your first step is incorrect. By stating $x^2=-\frac{36}{25}$ , you're adding another answer in which is why you have the $\pm$

Try this:
$\sqrt{-\frac{36}{25}}$
$=\sqrt{-1} \times \sqrt{(\frac{6}{5}) ^{2}}$
$=i \times \frac{6}{5}$
$=\frac{6}{5}i$

I would be very careful factoring that square root of -1

$1=\sqrt{1} = \sqrt{-1\times -1} = \sqrt{-1}\times \sqrt{-1} = -1$

as you can see, it's not exactly the 'right' thing to do in surds

what 'should' be done is $\sqrt{-\frac{36}{25}} = \sqrt{i^2 \times \frac{36}{25}} = \frac{6}{5}i$
so this would be the correct way to do it?
$\sqrt{-1}\times \sqrt{-1}$

so instead of putting
$\sqrt{-1}$ you'd put $\sqrt{-1}\times \sqrt{-1}$ and then expand to make it equal -1?
Title: Re: hards methods i mean specialist :D questions
Post by: shinny on November 19, 2008, 03:24:03 pm: Quote from: Mao on November 19, 2008, 03:07:02 pm
Quote from: shinjitsuzx on November 18, 2008, 09:45:13 pm
Your first step is incorrect. By stating $x^2=-\frac{36}{25}$ , you're adding another answer in which is why you have the $\pm$

Try this:
$\sqrt{-\frac{36}{25}}$
$=\sqrt{-1} \times \sqrt{(\frac{6}{5}) ^{2}}$
$=i \times \frac{6}{5}$
$=\frac{6}{5}i$

I would be very careful factoring that square root of -1

$1=\sqrt{1} = \sqrt{-1\times -1} = \sqrt{-1}\times \sqrt{-1} = -1$

as you can see, it's not exactly the 'right' thing to do in surds

what 'should' be done is $\sqrt{-\frac{36}{25}} = \sqrt{i^2 \times \frac{36}{25}} = \frac{6}{5}i$

Ah right didn't think of it that way. Normally I wouldn't write it like that but it seemed to explain things clearer in doing so.

Quote from: hard on November 19, 2008, 03:11:43 pm
so this would be the correct way to do it?
$\sqrt{-1}\times \sqrt{-1}$

so instead of putting
$\sqrt{-1}$ you'd put $\sqrt{-1}\times \sqrt{-1}$ and then expand to make it equal -1?

Just do exactly what Mao did. Don't split it up, but rather convert the negative (implied -1) inside the original surd, into $i^{2}$
Title: Re: hards methods i mean specialist :D questions
Post by: hard on November 19, 2008, 10:46:54 pm: okay the question asks:

Evaluate $3$ x $Re(1+i-i$ ³ $/i)$ + 2 x $Im(i$ ⁴ + $3i$ + $4/2i)$

So far this is what i have done:

$3$ x $Re(1+i-i$ ³ $/i)$ + 2 x $Im(i$ ⁴ + $3i$ + $4/2i)$

= $3$ x $(1+i-(i$ ²)i $/i)$ + 2 x $(i$ ²)² + $3i$ + $4/2i)$

= $3$ x $(1+i+i$ $/i)$ + 2 x $($ 1 + $3i$ + $4/2i)$

=( $3$ + $6i$ $/i$ ) + ( $2$ + $6i$ + $8$ $/2i)$ )

= $6$ + $12i$ + $2$ + $6i$ + $8$ $/2i$

= $16$ + $18i$ / $2i$

okay so after this i'm stuck; but not 100% sure i did it right.

the asnwer says $4-i$ but not sure how they got that?

can someone shed some light.
Title: Re: hards methods i mean specialist :D questions
Post by: Glockmeister on November 19, 2008, 10:52:53 pm: Is the Re and Im in the question. cause if it is, then there will be no imaginary number.
Title: Re: hards methods i mean specialist :D questions
Post by: hard on November 19, 2008, 10:57:05 pm: Quote from: Glockmeister on November 19, 2008, 10:52:53 pm
Is the Re and Im in the question. cause if it is, then there will be no imaginary number.

yer it's part of the question but what do you mean there will be no imaginary number. How so?
Title: Re: hards methods i mean specialist :D questions
Post by: shinny on November 19, 2008, 11:01:31 pm: Because you remove the 'i' when you take the imaginary part of something. e.g. $Im(1+2i)=2$ and in that case, the book's answer is definitely wrong.
Title: Re: hards methods i mean specialist :D questions
Post by: hard on November 19, 2008, 11:11:45 pm: Quote from: shinjitsuzx on November 19, 2008, 11:01:31 pm
Because you remove the 'i' when you take the imaginary part of something. e.g. $Im(1+2i)=2$ and in that case, the book's answer is definitely wrong.

so how would you do the question
Title: Re: hards methods i mean specialist :D questions
Post by: hard on November 19, 2008, 11:21:32 pm: ?????
Title: Re: hards methods i mean specialist :D questions
Post by: Glockmeister on November 19, 2008, 11:35:36 pm: When you say Re(z), you are basically saying that this is the co-efficient of the 'real' part of a complex number z. In the same manner, Im(z) refers to the co-efficient of the 'imaginary' part of a complex number z.
Title: Re: hards methods i mean specialist :D questions
Post by: shinny on November 19, 2008, 11:57:11 pm: Due to cbf, I'll just show the main steps and skip the algebra hacking.

$3 \times Re( \frac{1+i-i^{3}}{i})+2 \times Im(\frac{i^{4}+3i+4}{2i})$

$=3 \times Re(2-i) + 2 \times Im(\frac{3}{2}-\frac{5}{2}i)$

$=3 \times 2 + 2 \times \frac{-5}{2}$

$=6-5$

$=1$

To do the dividing in the first few steps, just multiply top and bottom by i to 'realise' the denominator.
Title: Re: hards methods i mean specialist :D questions
Post by: hard on November 20, 2008, 12:46:17 am: Quote from: shinjitsuzx on November 19, 2008, 11:57:11 pm
Due to cbf, I'll just show the main steps and skip the algebra hacking.

$3 \times Re( \frac{1+i-i^{3}}{i})+2 \times Im(\frac{i^{4}+3i+4}{2i})$

$=3 \times Re(2-i) + 2 \times Im(\frac{3}{2}-\frac{5}{2}i)$

$=3 \times 2 + 2 \times \frac{-5}{2}$

$=6-5$

$=1$

To do the dividing in the first few steps, just multiply top and bottom by i to 'realise' the denominator.

thanx