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fp1 complex numbers

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Original post by Zacken
Uh, I said equate the real and imaginary parts, not the brackets. If I write z=(1)(2)z = (1)(2) do you automatically assume that 1=ℜz1 = \Re z and 2=ℑz2 = \Im z? Write zz in the form x+iyx +iy, see if you can move on from there.

---

If you'll excuse me, some (harshly worded) advice:

Spoiler


I've never seen a question like that before.
I should probably give myself a bit more time to learn the stuff though and think the question through a bit more.
Original post by Chittesh14
Tan pi/4 is 1 :P. But anyway, no need to learn more trig this is in radians anyway lol. Now that you've got the expanded bracket correct. So, the number on the top and bottom are the same.

So, p - 3q = 3p + q.

Try to get that rearranged into the form the question is asking now :smile:.


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ok thanks
Original post by Zacken
This is basic trig, you don't need to learn "more" trig, you just need to learn basic trig... it may help if I tell you that arctan⁡\arctan is just another way of writing tan⁡−1\tan^{-1}. Also, tan⁡π2≠1\tan \frac{\pi}{2} \neq 1, so I'm not sure why you "know" it. In fact, the tangent of π2\frac{\pi}{2} is undefined, so that's literally one of the worst things you could have claimed to be equal to 1, since it's about as far off being equal to 1 as one can possible even get.


Anyways, to address your question: look at an argand diagram, you want the argument of a complex number to be pi/4, this means it needs to be a line that angled at pi/4 to the (positive) real axis and pi/4 to the (positive) imaginary axis. This is precisely (if you sketch the line) the line y=xy=x, i.e: ℜ(z)=ℑ(z)\Re (z) = \Im(z).


Oops sorry i got mixed up. Tan 90 doesn't have any values >.> big mistake from me.

Initially i did draw a diagram and realised it was an isosceles triangle but i didn't know where to go from there.
Original post by kinderbar
I've never seen a question like that before.
I should probably give myself a bit more time to learn the stuff though and think the question through a bit more.

ok thanks


Oops sorry i got mixed up. Tan 90 doesn't have any values >.> big mistake from me.

Initially i did draw a diagram and realised it was an isosceles triangle but i didn't know where to go from there.


Don't let him put you down lol. He is harsh with his words, but he's a nice guy otherwise. He just wants you to be sure of what you're saying and doing. If you're having problems sketching, just do what I said, you'll get the answer within a minute...

Rearrange this:

So, p - 3q = 3p + q.

In terms of the diagram, if it's isosceles then both sides are equal. So, again it leads to the equation above.

Now, rearrange that equation so you get p + 2q = 0

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(edited 7 years ago)
Original post by physicsmaths
Some real misconceptions in here.
arg is just an angle, you find with tan^-1 of img/R.


just realised this is the reverse of a question asking you to find the mod/arg of a complex number -.-'
Original post by kinderbar
just realised this is the reverse of a question asking you to find the mod/arg of a complex number -.-'


How far are you into FP1?


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Original post by Chittesh14
How far are you into FP1?


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supposedly only finished the first chapter
btw fo the answer to part b i got q as 25\sqrt \dfrac{\sqrt 2}{5} is that right?
Original post by kinderbar
supposedly only finished the first chapter
btw fo the answer to part b i got q as 25\sqrt \dfrac{\sqrt 2}{5} is that right?


Nope :/ I kept making silly mistakes on that getting stupid answers like that too.
Let me give you a hint.

I'm assuming you've worked out that the mod of z = the square root of 10p^2 + 10q^2, which in turn is equal to 10root2.

Now, square both sides to get 200 = 10p^2 + 10q^2.
Now, use the second simultaneous equation p + 2q = 0 to solve for p and q.

I used substitution to solve for p and q.


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Original post by Chittesh14
Nope :/ I kept making silly mistakes on that getting stupid answers like that too.
Let me give you a hint.

I'm assuming you've worked out that the mod of z = the square root of 10p^2 + 10q^2, which in turn is equal to 10root2.

Now, square both sides to get 200 = 10p^2 + 10q^2.
Now, use the second simultaneous equation p + 2q = 0 to solve for p and q.

I used substitution to solve for p and q.


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There's another way.
As ∣z∣=102 |z|=10\sqrt 2 and arg z=π/4 \text{arg z} = \pi /4 , you can say that 102cos⁡(π/4)=p−3q 10\sqrt 2 \cos (\pi /4)=p-3q and 102sin⁡(π/4)=3p+q 10\sqrt 2 \sin (\pi /4)=3p+q . Then it's easy linear simultaneous equations.
Original post by Chittesh14
Nope :/ I kept making silly mistakes on that getting stupid answers like that too.
Let me give you a hint.

I'm assuming you've worked out that the mod of z = the square root of 10p^2 + 10q^2, which in turn is equal to 10root2.

Now, square both sides to get 200 = 10p^2 + 10q^2.
Now, use the second simultaneous equation p + 2q = 0 to solve for p and q.

I used substitution to solve for p and q.


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uh i got

10p^2 +10q^2 =10root2

from simplifying from (3p+q)^2 + (p-3q)^2 =10root2
Eh... Do you understand what you are doing?
Original post by kinderbar
uh i got

10p^2 +10q^2 =10root2

from simplifying from (3p+q)^2 + (p-3q)^2 =10root2


It should be 10 root 2 = square root of 10p^2 + 10q^2 lol.
Remember, the mod is square root of z^2 so it is the square root of what you simplified.


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Original post by B_9710
There's another way.
As ∣z∣=102 |z|=10\sqrt 2 and arg z=π/4 \text{arg z} = \pi /4 , you can say that 102cos⁡(π/4)=p−3q 10\sqrt 2 \cos (\pi /4)=p-3q and 102sin⁡(π/4)=3p+q 10\sqrt 2 \sin (\pi /4)=3p+q . Then it's easy linear simultaneous equations.


Oh wow, mod-arg form. That's such a sexy method. Dam man, you're a genius lol.


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Original post by Imperion
Eh... Do you understand what you are doing?


Original post by Chittesh14
It should be 10 root 2 = square root of 10p^2 + 10q^2 lol.
Remember, the mod is square root of z^2 so it is the square root of what you simplified.


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*facepalm my stupidity knows no bounds

so now i got q as 2 and p=-4
(edited 7 years ago)
Original post by kinderbar
*facepalm my stupidity knows no bounds

so now i got q as 2 and p=-4


You got the minus signs the wrong way round.
Original post by kinderbar
*facepalm my stupidity knows no bounds

so now i got q as 2 and p=-4


Yeah, other way round. Should be q = -2 and p = 4.


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Original post by B_9710
You got the minus signs the wrong way round.


Original post by Chittesh14
Yeah, other way round. Should be q = -2 and p = 4.


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But q could be plus or minus 2?
Original post by kinderbar
But q could be plus or minus 2?


You have to decide which one it is.
Original post by B_9710
You have to decide which one it is.


both options of q= +or- 2 and p= +or- 4 work in the equation p+2q=0 and in the equation
Unparseable latex formula:

\dfrac{3p+q]{p-3q}



....

good thing i read the question -.-'

so the arg of z is just pi/4 ?
Original post by kinderbar
both options of q= +or- 2 and p= +or- 4 work in the equation p+2q=0 and in the equation
Unparseable latex formula:

\dfrac{3p+q]{p-3q}



....

good thing i read the question -.-'

so the arg of z is just pi/4 ?


Exactly arg z=π/4 \text{arg z}=\pi /4 so there is only one set of values of p and q which work. Use your different values and look at what you get for your value of z z and you should see which is correct.
Or just simply forget all that. If p > 0 as it says in the question, then p is obviously 4 and q is -2 lo.l


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Original post by B_9710
Exactly arg z=π/4 \text{arg z}=\pi /4 so there is only one set of values of p and q which work. Use your different values and look at what you get for your value of z z and you should see which is correct.


it just said in the question that p>0 so i know that q=-2 :smile:

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