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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) do you automatically assume that 1 = \Re z and 2 = \Im z? Write z in the form x +iy, see if you can move on from there.

    ---

    If you'll excuse me, some (harshly worded) advice:
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    It sounds like you're more simply learning/memorising algorithms to answer questions without having a single bit of understanding of the actual mathematics or indeed much of a clue as to what you're actually doing beyond imitating examples or doing random things like equating brackets and hoping the right answer falls out, which, if it promptly does, you kid yourself into believing that you understand it.

    You're kidding yourself if you think you're giving yourself an advantage by "learning" this content early since you're not really learning anything and indeed, this is just going to be detrimental for you since it'll stunt your understanding. So either buckle up and learn things properly or stop trying this farce of teaching yourself.
    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 .


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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 is just another way of writing \tan^{-1}. Also, \tan \frac{\pi}{2} \neq 1, so I'm not sure why you "know" it. In fact, the tangent of \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=x, i.e: \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.
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    (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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    (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 -.-'
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    (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 \sqrt \dfrac{\sqrt 2}{5} is that right?
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    (Original post by kinderbar)
    supposedly only finished the first chapter
    btw fo the answer to part b i got q as \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|=10\sqrt 2 and  \text{arg z} = \pi /4 , you can say that  10\sqrt 2 \cos (\pi /4)=p-3q and  10\sqrt 2 \sin (\pi /4)=3p+q . Then it's easy linear simultaneous equations.
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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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    uh i got

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

    from simplifying from (3p+q)^2 + (p-3q)^2 =10root2
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    Eh... Do you understand what you are doing?
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    (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|=10\sqrt 2 and  \text{arg z} = \pi /4 , you can say that  10\sqrt 2 \cos (\pi /4)=p-3q and  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
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    (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.
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    (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?
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    (Original post by kinderbar)
    But q could be plus or minus 2?
    You have to decide which one it is.
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    (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 or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    \dfrac{3p+q]{p-3q}


    ....

    good thing i read the question -.-'

    so the arg of z is just pi/4 ?
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    (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 or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    \dfrac{3p+q]{p-3q}


    ....

    good thing i read the question -.-'

    so the arg of z is just pi/4 ?
    Exactly  \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 and you should see which is correct.
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    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  \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 and you should see which is correct.
    it just said in the question that p>0 so i know that q=-2
 
 
 
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