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    I'm doing a question but am stuck on one bit.

    Z1= P+2i
    Z2= 1-2i

    Part (a) says find Z1/Z2.

    I got

    (P+2i)/(1-2i) x (1+2i)/(1+2i) = (P-4)/5 +(2P+2)I/5

    Part (b): |z1/z2| = 13, find the possible values of P

    So, Done a long way but ended up with the correct answer (+/- 29). And I went on google and wrote 'FP1 June 2014 answers' and I found Arsey's model answers.

    He done the same thing as me for part A, but for part B he done 2 different methods, the one i done and another one.
    It was

    (p^2 + 4)/(1 + 4) = 169
    (P^2 + 4) = 845
    P^2 = 841
    P = +/- 29

    Now this way is much quicker, but i don't understand where the (p^2 + 4) came from?

    Can someone please tell me where he got it from?
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    (Original post by Economistician)
    I'm doing a question but am stuck on one bit.

    Z1= P+2i
    Z2= 1-2i

    Part (a) says find Z1/Z2.

    I got

    (P+2i)/(1-2i) x (1+2i)/(1+2i) = (P-4)/5 +(2P+2)I/5

    Part (b): |z1/z2| = 13, find the possible values of P

    So, Done a long way but ended up with the correct answer (+/- 29). And I went on google and wrote 'FP1 June 2014 answers' and I found Arsey's model answers.

    He done the same thing as me for part A, but for part B he done 2 different methods, the one i done and another one.
    It was

    (p^2 + 4)/(1 + 4) = 169
    (P^2 + 4) = 845
    P^2 = 841
    P = +/- 29

    Now this way is much quicker, but i don't understand where the (p^2 + 4) came from?

    Can someone please tell me where he got it from?
    (a+bi)(a-bi) = a2 + b2
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    (Original post by TeeEm)
    (a+bi)(a-bi) = a2 + b2
    Sorry, I don't understand. I see how that works in essence, but how did he know he had to use p+2i and p-2i?
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    (Original post by Economistician)
    Sorry, I don't understand. I see how that works in essence, but how did he know he had to use p+2i and p-2i?
    I have not read the question but simply he "conjugated" the denominator in order to process the algebra.
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    (Original post by TeeEm)
    I have not read the question but simply he "conjugated" the denominator in order to process the algebra.
    http://www.physicsandmathstutor.com/...-maths-papers/
    its the first question on the one that says "Edexcel FP1 - June 2014 Model Answers by Arsey.pdf"
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    correction to what I said previously as I could not see the question.

    |z1/z2| =|z1|/|z2|

    now
    the modulus of z1 is the square root of of p2 + 22 = p2+4
    the modulus of z2 is the square root of of 12 + (-2)2 = 5

    squaring the fraction = 132

    ....
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    (Original post by TeeEm)
    correction to what I said previously as I could not see the question.

    |z1/z2| =|z1|/|z2|

    now
    the modulus of z1 is the square root of of p2 + 22 = p2+4
    the modulus of z2 is the square root of of 12 + (-2)2 = 5

    squaring the fraction = 132
    ....
    Thank you, really appreciate it.
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    (Original post by Economistician)
    Thank you, really appreciate it.
    no worries
 
 
 
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