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FP1 further maths complex numbers question Watch

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    I don't understand why |z+3i| has the cartesian equation x^2+(y-3)^2=9 and |z-4i|=5 has the C.E. x^2+(y-4)^2=25. Why are both y components of the equation negative, so -3 and -4 when the modulus counterparts are of different signs, so not both negative or positive, but one is negative and one is positive. How can they have the same cartesian equation format but be different in the modulus sign? Is it because of the modulus sign, which makes everything inside positive? Thanks

    ^^ title is meant to say FP2
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    the statement | z + a + bi | = k

    corresponds to a circle with radius k and centre ( -a, -b )

    z - ( -a - bi ) can be thought of as a vector from ( -a -bi ) to z... so |z - ( -a - bi )| is th length of this vector.
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    (Original post by the bear)
    the statement | z + a + bi | = k

    corresponds to a circle with radius k and centre ( -a, -b )

    z - ( -a - bi ) can be thought of as a vector from ( -a -bi ) to z... so |z - ( -a - bi )| is th length of this vector.

    How would you write that in cartesian form? Because there doesn't seem to be a rule to follow each time for how to translate the modulus form into the cartesian in terms of -ve's and +ve's. I'm just doing the question |z-1-i| which solution bank says translates to (x-1)^2+(y-1)^2=25 but this does not follow the same pattern as |z-3| which is supposed to translate to (x+3)^2+y^2=4 but this doesn't follow any pattern, so how can this be? Because the -ve in the modulus sign does not consistently translate to a -ve in the cartesian equation, unless solution bank made a mistake. How do we know when to use the -ve or +ve for the cartesian equation?
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    (Original post by adriannala)
    How would you write that in cartesian form? Because there doesn't seem to be a rule to follow each time for how to translate the modulus form into the cartesian in terms of -ve's and +ve's. I'm just doing the question |z-1-i| which solution bank says translates to (x-1)^2+(y-1)^2=25 but this does not follow the same pattern as |z-3| which is supposed to translate to (x+3)^2+y^2=4 but this doesn't follow any pattern, so how can this be? Because the -ve in the modulus sign does not consistently translate to a -ve in the cartesian equation, unless solution bank made a mistake. How do we know when to use the -ve or +ve for the cartesian equation?
    |z-1-i| can be thought of as the length of a vector from ( 1, 1 ) to z

    so if |z-1-i| = 5 that will be a circle of centre ( 1, 1 ) and radius 5... so

    ( x - 1 )2 + ( y - 1 )2 = 25
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    (Original post by the bear)
    |z-1-i| can be thought of as the length of a vector from ( 1, 1 ) to z

    so if |z-1-i| = 5 that will be a circle of centre ( 1, 1 ) and radius 5... so

    ( x - 1 )2 + ( y - 1 )2 = 25
    I still don't understand why |z+3i|=3 goes to a -3 in the cartesian equation but |z-4i|=5 also goes to a -4 in the cartesian equation?? Surely there is some t=kind of swap over as in the +ve goes to the -ve? otherwise there is no rule? So how do we know when to use the +ve or the -ve for the cartesian equation? Why does |z+1|=1 go to (x-1)^2 etc in cartesian form but |z-3|=2 go to a +ve in cartesian form? thanks
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    (Original post by adriannala)
    I still don't understand why |z+3i|=3 goes to a -3 in the cartesian equation but |z-4i|=5 also goes to a -4 in the cartesian equation?? Surely there is some t=kind of swap over as in the +ve goes to the -ve? otherwise there is no rule? So how do we know when to use the +ve or the -ve for the cartesian equation? Why does |z+1|=1 go to (x-1)^2 etc in cartesian form but |z-3|=2 go to a +ve in cartesian form? thanks
    Huh?

    |z-(a+b\mathbf{i})|=r in the complex plane means (x-a)^2+(y-b)^2=r^2 in the xy plane.

    So |z+3\mathbf{i}|=3 goes to x^2+(y+3)^2=9

    and |z-4\mathbf{i}|= 5 goes to x^2+(y-4)^2=25
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    (Original post by adriannala)
    I still don't understand why |z+3i|=3 goes to a -3 in the cartesian equation but |z-4i|=5 also goes to a -4 in the cartesian equation??
    that does sound if there is a mistake somewhere...


    |z+3i|=3 means " the length of a vector from ( 0, -3 ) to z is 3 "

    |z-4i|=5 means " the length of a vector from ( 0, 4 ) to z is 5 "
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    (Original post by RDKGames)
    Huh?

    |z-(a+b\mathbf{i})|=r in the complex plane means (x-a)^2+(y-b)^2=r^2 in the xy plane.

    So |z+3\mathbf{i}|=3 goes to x^2+(y-3)^2=9

    and |z-4\mathbf{i}|= 5 goes to x^2+(y+4)^2=25 - it doesn't have a -4 in there, not sure where you saw this...
    I think solution bank made a mistake as that's where I was checking my answers.
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    (Original post by RDKGames)
    Huh?

    |z-(a+b\mathbf{i})|=r in the complex plane means (x-a)^2+(y-b)^2=r^2 in the xy plane.

    So |z+3\mathbf{i}|=3 goes to x^2+(y-3)^2=9

    and |z-4\mathbf{i}|= 5 goes to x^2+(y+4)^2=25 - it doesn't have a -4 in there, not sure where you saw this...
    Thanks for clearing that up.
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    you're welcome.
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    (Original post by adriannala)
    I think solution bank made a mistake as that's where I was checking my answers.
    What is the solution bank that you used?
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    (Original post by RDKGames)
    What is the solution bank that you used?
    Physics and maths solution bank, do you know it?
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    (Original post by adriannala)
    Physics and maths solution bank, do you know it?
    Yeah if it's for the Heinemann textbooks then those solution banks have plenty of mistakes across all modules as I've seen many other students come to the forum with them, so don't be surprised if you find another.
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    (Original post by RDKGames)
    Yeah if it's for the Heinemann textbooks then those solution banks have plenty of mistakes across all modules as I've seen many other students come to the forum with them, so don't be surprised if you find another.
    Oh ok thanks for letting me know.
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    (Original post by RDKGames)
    Huh?

    |z-(a+b\mathbf{i})|=r in the complex plane means (x-a)^2+(y-b)^2=r^2 in the xy plane.

    So |z+3\mathbf{i}|=3 goes to x^2+(y-3)^2=9

    and |z-4\mathbf{i}|= 5 goes to x^2+(y+4)^2=25 - it doesn't have a -4 in there, not sure where you saw this...
    I think the equations of the circles are actually x^2+(y+3)^3=9 and
    x^2+(y-4)^2=25, isn't it?
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    (Original post by danmorris134)
    I think the equations of the circles are actually x^2+(y+3)^3=9 and
    x^2+(y-4)^2=25, isn't it?
    Yeah you're right, small error there, amended.

    (Original post by adriannala)
    Oh ok thanks for letting me know.
    Note the above error I've made. In either case, there is still an error in the solution bank, as there is no -3 instead.
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    How do you sketch the locus of |z+3|=3|z-5|?
    And what is the cartesian equation of this? But mostly how do you get to the answer?
    Thanks
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    (Original post by adriannala)
    How do you sketch the locus of |z+3|=3|z-5|?
    And what is the cartesian equation of this? But mostly how do you get to the answer?
    Thanks
    Write Z as X+iY then collect real and imaginary terms. You know that the modulus of a complex number z=X+iY is root(X^2 + Y^2). Then square both sides and rearrange into equation of a circle.
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    (Original post by adriannala)
    How do you sketch the locus of |z+3|=3|z-5|?
    And what is the cartesian equation of this? But mostly how do you get to the answer?
    Thanks
    It's easier here to find the Cartesian equation first, as it will still represent the locus of points satisfying that equality, while giving a more intuitive idea of how it looks like.

    Begin by saying that |(x+3)+yi|=3|(x-5)+yi| then use the fact that |z|=\sqrt{x^2+y^2} for z=x+iy

    So here we just have \sqrt{(x+3)^2+y^2}=3\sqrt{(x-5)^2+y^2}

    Then just work the algebra to get a nice circle of loci, then sketch it.
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    Another question, if we are given that |z-2-2i|=2 and arg(z-2-2i)=pi/6 how do we find the value for z in the form a+ib? Thanks.
 
 
 
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