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    (Original post by timebent)
    how does one do part A???
    arg(z) = pi/4 if the imaginary and real parts of z are equal (sketch...), and you're pretty much done from then.
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    (Original post by timebent)
    how late u getting up xD
    Usually get up at 2 p.m., but today is your lucky day I guess. I'm out of bed but I have to eat. I can see Zacken has done some explanation, try that and if you still get it wrong or feel your wrong - post your workings and I'll take a look,


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    (Original post by Zacken)
    arg(z) = pi/4 if the imaginary and real parts of z are equal (sketch...), and you're pretty much done from then.
    i don't quite understand, if i equate the brackets of z then i get p as 1 and q as 3??? I'm most certainly doing something wrong but i don;t know what >.>
    (Original post by Chittesh14)
    Usually get up at 2 p.m., but today is your lucky day I guess. I'm out of bed but I have to eat. I can see Zacken has done some explanation, try that and if you still get it wrong or feel your wrong - post your workings and I'll take a look,


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    That's late haha.
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    (Original post by kinderbar)
    i don't quite understand, if i equate the brackets of z then i get p as 1 and q as 3??? I'm most certainly doing something wrong but i don;t know what >.>

    That's late haha.
    I'll brush my teeth and after 10 minutes I'll help you. Try to post a picture of your working out till then .


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    (Original post by kinderbar)
    i don't quite understand, if i equate the brackets of z then i get p as 1 and q as 3??? I'm most certainly doing something wrong but i don;t know what .
    If  \text{arg(z)}=\pi /4 , the the real part of z and the imaginary part of z are equal. Did you get z to be what I had it yesterday?
    (Original post by B_9710)
    It's just simultaneous equations. You have  z= p-3q +(3p+q)i and  p+2q=0 .
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    (Original post by B_9710)
    If  \text{arg(z)}=\pi /4 , the the real part of z and the imaginary part of z are equal. Did you get z to be what I had it yesterday?
    No, the answers don't match :/. But, I guess he is on the right track because he's got integers as his answers - but something must've gone wrong.


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    I've solved it.

    I don't understand how square root (10p^2 + 10q^2) doesn't equal to root10 * p + root 10 * q lol... That caused me mistakes about 3 times .


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    (Original post by Chittesh14)
    I'll brush my teeth and after 10 minutes I'll help you. Try to post a picture of your working out till then .


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    i have no working ;_;
    (Original post by B_9710)
    If  \text{arg(z)}=\pi /4 , the the real part of z and the imaginary part of z are equal. Did you get z to be what I had it yesterday?
    no i didn't get what you got :/

    where do all those numbers come from??? o.o
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    (Original post by kinderbar)
    i have no working ;_;


    no i didn't get what you got :/

    where do all those numbers come from??? o.o
    Well you have  z=(1+3i)(p+qi) , you just have to expand the brackets.
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    (Original post by kinderbar)
    i have no working ;_;


    no i didn't get what you got :/

    where do all those numbers come from??? o.o
    Ok let me explain. If arg z = pi / 4, then clearly it's indicating that arctan (y/x) is 1. So, you know the y and the x are equal.
    So what you do is expand z and then rewrite it in the form a + bi.
    Then do b/a where b represents y and a represents x.
    But, since they divide to give 1 as we know, b = x.

    Eventually, after rearranging it you will get p + 2q = 0


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    (Original post by kinderbar)
    i don't quite understand, if i equate the brackets of z then i get p as 1 and q as 3??? I'm most certainly doing something wrong but i don;t know what >.>
    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:

    Spoiler:
    Show
    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.
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    (Original post by B_9710)
    Well you have  z=(1+3i)(p+qi) , you just have to expand the brackets.
    all expanded to give what you got
    (p-3q)+(3p+q)i
    (Original post by Chittesh14)
    Ok let me explain. If arg z = pi / 4, then clearly it's indicating that arctan (y/x) is 1. So, you know the y and the x are equal.
    So what you do is expand z and then rewrite it in the form a + bi.
    Then do b/a where b represents y and a represents x.
    But, since they divide to give 1 as we know, b = x.

    Eventually, after rearranging it you will get p + 2q = 0


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    Should i learn some more trig? i don't really know what arc tand is but i do know that tan pi/2 =1 so the number on the top and bottom are the same
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    (Original post by Chittesh14)
    I've solved it.

    I don't understand how square root (10p^2 + 10q^2) doesn't equal to root10 * p + root 10 * q lol... That caused me mistakes about 3 times .


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    Of course it doesn't... do you think that \sqrt{1 + 1} = \sqrt{1} + \sqrt{1} = 1 + 1 = 2 \Rightarrow \sqrt{2} = 2...? i.e: you think that the square root of 2 is 2, so you need to multiply 2 by itself twice to get 2?

    i.e: you seem to be under the impression that \sqrt{a+b} = \sqrt{a} + \sqrt{b} when the teeniest bit of checking would reveal that that is not even close to an identity. I've written a post about the linearity of the square root previously, see if you can find it on my profile, if you want some intuition about it.
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    (Original post by kinderbar)
    all expanded to give what you got
    (p-3q)+(3p+q)i

    Should i learn some more trig? i don't really know what arc tand is but i do know that tan pi/2 =1 so the number on the top and bottom are the same
    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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    (Original post by Zacken)
    Of course it doesn't... do you think that \sqrt{1 + 1} = \sqrt{1} + \sqrt{1} = 1 + 1 = 2 \Rightarrow \sqrt{2} = 2...? i.e: you think that the square root of 2 is 2, so you need to multiply 2 by itself twice to get 2?

    i.e: you seem to be under the impression that \sqrt{a+b} = \sqrt{a} + \sqrt{b} when the teeniest bit of checking would reveal that that is not even close to an identity. I've written a post about the linearity of the square root previously, see if you can find it on my profile, if you want some intuition about it.
    I'm fine thanks.... I am a bit stupid so I got that wrong clearly. I realised after, but still didn't understand it. Maybe I should've substituted numbers into p and q to test it. Thanks for the explanation lolz.


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    (Original post by kinderbar)
    Should i learn some more trig? i don't really know what arc tand is but i do know that tan pi/2 =1 so the number on the top and bottom are the same
    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).

    (Original post by Chittesh14)
    Tan pi/4 is 1 :P. But anyway, no need to learn more trig this is in radians anyway lol.
    Trig is in radians... obviously. How does "this is radians" imply that it isn't trig? That's as contradictory as you get.
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    (Original post by Chittesh14)
    I realised after, but still didn't understand it.
    If you want to understand it, then try and find my post about the linearity on the square root. I'll look for it myself later, no time right now. But I'm not sure you really care about understanding it anyway, so meh.
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    (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).



    Trig is in radians... obviously. How does "this is radians" imply that it isn't trig? That's as contradictory as you get.
    Stop being so harsh lol.

    And when did I say it isn't trig?


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    (Original post by Zacken)
    If you want to understand it, then try and find my post about the linearity on the square root. I'll look for it myself later, no time right now. But I'm not sure you really care about understanding it anyway, so meh.
    Let's see, I might if I have time myself later.


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    Some real misconceptions in here.
    arg is just an angle, you find with tan^-1 of img/R.
 
 
 
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