I don't understand why z+3i has the cartesian equation x^2+(y3)^2=9 and z4i=5 has the C.E. x^2+(y4)^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

adriannala
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 11112017 10:10

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the bear
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 11112017 10:20
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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 11112017 10:28
(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 z1i which solution bank says translates to (x1)^2+(y1)^2=25 but this does not follow the same pattern as z3 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? 
the bear
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 11112017 10:31
(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 z1i which solution bank says translates to (x1)^2+(y1)^2=25 but this does not follow the same pattern as z3 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?
so if z1i = 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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 11112017 10:48
(Original post by the bear)
z1i can be thought of as the length of a vector from ( 1, 1 ) to z
so if z1i = 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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 11112017 10:52
(Original post by adriannala)
I still don't understand why z+3i=3 goes to a 3 in the cartesian equation but z4i=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 (x1)^2 etc in cartesian form but z3=2 go to a +ve in cartesian form? thanks
in the complex plane means in the xy plane.
So goes to
and goes toLast edited by RDKGames; 11112017 at 11:59. 
the bear
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 11112017 10:52
(Original post by adriannala)
I still don't understand why z+3i=3 goes to a 3 in the cartesian equation but z4i=5 also goes to a 4 in the cartesian equation??
z+3i=3 means " the length of a vector from ( 0, 3 ) to z is 3 "
z4i=5 means " the length of a vector from ( 0, 4 ) to z is 5 " 
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 11112017 11:06
(Original post by RDKGames)
Huh?
in the complex plane means in the xy plane.
So goes to
and goes to  it doesn't have a 4 in there, not sure where you saw this... 
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 11112017 11:06
(Original post by RDKGames)
Huh?
in the complex plane means in the xy plane.
So goes to
and goes to  it doesn't have a 4 in there, not sure where you saw this... 
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 11112017 11:07
you're welcome.

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 11112017 11:10
(Original post by adriannala)
I think solution bank made a mistake as that's where I was checking my answers. 
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 11112017 11:16
(Original post by RDKGames)
What is the solution bank that you used? 
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 11112017 11:19
(Original post by adriannala)
Physics and maths solution bank, do you know it? 
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 11112017 11:19
(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. 
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 11112017 11:50
(Original post by RDKGames)
Huh?
in the complex plane means in the xy plane.
So goes to
and goes to  it doesn't have a 4 in there, not sure where you saw this...
x^2+(y4)^2=25, isn't it? 
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 11112017 12:00
(Original post by danmorris134)
I think the equations of the circles are actually x^2+(y+3)^3=9 and
x^2+(y4)^2=25, isn't it?
(Original post by adriannala)
Oh ok thanks for letting me know. 
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 11112017 13:31
How do you sketch the locus of z+3=3z5?
And what is the cartesian equation of this? But mostly how do you get to the answer?
Thanks 
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 11112017 13:54
(Original post by adriannala)
How do you sketch the locus of z+3=3z5?
And what is the cartesian equation of this? But mostly how do you get to the answer?
Thanks 
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 11112017 13:58
(Original post by adriannala)
How do you sketch the locus of z+3=3z5?
And what is the cartesian equation of this? But mostly how do you get to the answer?
Thanks
Begin by saying that then use the fact that for
So here we just have
Then just work the algebra to get a nice circle of loci, then sketch it. 
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 12112017 11:53
Another question, if we are given that z22i=2 and arg(z22i)=pi/6 how do we find the value for z in the form a+ib? Thanks.
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