# FP2: complex numbers

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#1
both complex numbers give me the same answer, for part c....
Last edited by Maths&physics; 1 year ago
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1 year ago
#2
I don't follow which part of the question you're trying to answer / what you're having trouble with?
Some bits of the answer (1st equation) don't make sense - the complex number does not equal its angle?
(Original post by Maths&physics)
both complex numbers give me the same answer.
Last edited by mqb2766; 1 year ago
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#3
(Original post by mqb2766)
I don't follow which part of the question you're trying to answer / what you're having trouble with?
Some bits of the answer (1st equation) don't make sense - the complex number does not equal its angle?
part c....

sorry, I rushed it a little to make it eligible for TSR....

it was arg ((x -6) +iy) = [-3(pi)]/4
Last edited by Maths&physics; 1 year ago
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1 year ago
#4
What did you get for a) and b)?
(Original post by Maths&physics)
part c....
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#5
(Original post by mqb2766)
What did you get for a) and b)?
I got them both right: centre: (4, -2) and radius (root)20

b was drawing the half line....
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1 year ago
#6
So c) is asking for the complex number which satisfies both a) and b). Should be easy to sketch, I'm presuming you just need to do the numbers to get the value?
Your working in the attached image seems very confused (argument incorrect, center 4,4, ...). What are you trying to do?
(Original post by Maths&physics)
I got them both right: centre: (4, -2) and radius (root)20

b was drawing the half line....
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1 year ago
#7
I'm not sure what you mean?

I got the same values of x and y for part c
X = 4 +/- sqrt 10
Y = -2 +/- sqrt 10
(Original post by Maths&physics)
both complex numbers give me the same answer, for part c....
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#8
(Original post by mqb2766)
So c) is asking for the complex number which satisfies both a) and b). Should be easy to sketch, I'm presuming you just need to do the numbers to get the value?
Your working in the attached image seems very confused (argument incorrect, center 4,4, ...). What are you trying to do?
when I subbed y = x - 6 into the equation of the circle, that's what you get and solving for x gave me the right x value but it gave me 2 and I didn't know which one was the right one. yeah, I am very confused by this question
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1 year ago
#9

Sorry if this is not what you're looking for
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#10
(Original post by ohemgee11)
I'm not sure what you mean?

I got the same values of x and y for part c
X = 4 +/- sqrt 10
Y = -2 +/- sqrt 10
so, I got cartesian form y = x - 6 from the second equation and subbed that into the circle equation and I got the right value for x..... but I got 2 values and I don't know which one is the right one and why...
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1 year ago
#11
Oh okay!
I am a bit rusty at Further maths at the moment but maybe they want you to combine the two results into one complex number?
But if they're looking for only one I can't remember how to find the 'correct' answer if there is a way to do so

Maybe others who are more advanced can help

(Original post by Maths&physics)
when I subbed y = x - 6 into the equation of the circle, that's what you get and solving for x gave me the right x value but it gave me 2 and I didn't know which one was the right one. yeah, I am very confused by this question
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1 year ago
#12
(Original post by Maths&physics)
so, I got cartesian form y = x - 6 from the second equation and subbed that into the circle equation and I got the right value for x..... but I got 2 values and I don't know which one is the right one and why...
Oh oops!
I think I found it

Of your two complex numbers, one is in the 1st quadrant, and the other is in the 3rd quadrant (just sketch them quickly on an argand diagram). The one in the first quadrant gives you an argument of pi/4
The one in the 3rd quadrant gives you -3pi/4. This is the one you're looking for

So z= 4 - sqrt 10 + (-2-sqrt 10)i

This is the complex number such that arg (z-6) = -3pi/4
Last edited by ohemgee11; 1 year ago
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#13
(Original post by ohemgee11)
Oh oops!
I think I found it

Of your two complex numbers, one is in the 1st quadrant, and the other is in the 3rd quadrant (just sketch them quickly on an argand diagram). The one in the first quadrant gives you an argument of pi/4
The one in the 3rd quadrant gives you -3pi/4. This is the one you're looking for

So z= 4 - sqrt 10 + (-2-sqrt 10)i

This is the complex number such that arg (z-6) = -3pi/4
when you solve this for theta, it doesn't give -3pi/4?
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1 year ago
#14
Did you subtract 6 from Re(z)?
(Original post by Maths&physics)
when you solve this for theta, it doesn't give -3pi/4?
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#15
what do you mean?

those values you said are the right answer (which they are) dont = -3pi/4 after you've drawn the diagram with those values.
(Original post by ohemgee11)
Did you subtract 6 from Re(z)?
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1 year ago
#16
Yea the -3pi/4 comes from arg(z - 6) not arg(z)

So in this case arg(z-6) =arg ((-2- sqrt 10) + (-2-sqrt 10)i)
This will give you -3pi/4

(Original post by Maths&physics)
what do you mean?

those values you said are the right answer (which they are) dont = -3pi/4 after you've drawn the diagram with those values.
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#17
(Original post by ohemgee11)

Sorry if this is not what you're looking for
that's what I got but there is only one answer, not 2.
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#18
(Original post by ohemgee11)
Yea the -3pi/4 comes from arg(z - 6) not arg(z)

So in this case arg(z-6) =arg ((-2- sqrt 10) + (-2-sqrt 10)i)
This will give you -3pi/4
why: ((-2- sqrt 10) + (-2-sqrt 10)i)

and not: ((-2 + sqrt 10) + (-2 + sqrt 10)i)
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1 year ago
#19
The argument of the second one is pi/4

The reason why you get two answers is because when you find the equation of the half line, you apply the tangent of -3pi/4, right?
Tan (-3pi/4) gives 1, but as you know, there is more than 1 angle that will also give tan (theta) is 1
So when we find x, we get two answers. These two answers will work for the circle equation, but only one will work for the half line, because only one has an argument of -3pi/4. The other has an argument of pi/4.
The one with an argument of -3pi/4 is the one you should use
(Original post by Maths&physics)
why: ((-2- sqrt 10) + (-2-sqrt 10)i)

and not: ((-2 + sqrt 10) + (-2 + sqrt 10)i)
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