# Complex numbers and their geometrical properties

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#1
An ant walks forward one unit and then turns to the right by 2pi/9. It repeats this a further 3 times. Show that the distance of the ant from its initial position is sin(4pi/0)/sin(pi/9).

Can someone show me their approach?
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2 years ago
#2
An ant walks forward one unit and then turns to the right by 2pi/9. It repeats this a further 3 times. Show that the distance of the ant from its initial position is sin(4pi/0)/sin(pi/9).

Can someone show me their approach?
It first walks 1 forward, then e^(2pi/9) then e^(4pi/9) then e^(6pi/9).
The position is given by the sum of all this.
Take its magnitude and you have the distance.
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#3
(Original post by golgiapparatus31)
It first walks 1 forward, then e^(2pi/9) then e^(4pi/9) then e^(6pi/9).
The position is given by the sum of all this.
Take its magnitude and you have the distance.
So you just add all of the complex numbers? I can't imagine how this would work.
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2 years ago
#4
So you just add all of the complex numbers? I can't imagine how this would work.
It is vectors
The real part gives the x-displacement
The imaginary gives the y-displacement

the complex numbers behave like vectors

if you don't believe me then try it

say you go east by 2 then north by 1
your position is described by 2+i
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1 week ago
#5
(Original post by golgiapparatus31)
It is vectors
The real part gives the x-displacement
The imaginary gives the y-displacement

the complex numbers behave like vectors

if you don't believe me then try it

say you go east by 2 then north by 1
your position is described by 2+i
i know this is really late lol but would that just be (e^2pi/9)^4?
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1 week ago
#6
(Original post by Htx_x346)
i know this is really late lol but would that just be (e^2pi/9)^4?
No. That is 4 equal rotations without any stepping.
The problem is a unit step, then rotate, then unit step, then rotate.
Last edited by mqb2766; 1 week ago
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1 week ago
#7
(Original post by mqb2766)
No. That is 4 equal rotations without any stepping.
The problem is a unit step, then rotate, then unit step, then rotate.
oh. hmmm.
but the magnitude is 1?

so would I have to add e^2pi/9 4 times?
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1 week ago
#8
(Original post by Htx_x346)
oh. hmmm.
but the magnitude is 1?

so would I have to add e^2pi/9 4 times?
Using complex numbers just makes the problem more complex that it needs to. You could get the answer by drawing a diagram and using trig
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1 week ago
#9
(Original post by Htx_x346)
oh. hmmm.
but the magnitude is 1?

so would I have to add e^2pi/9 4 times?
For the first point, you've just done demoivre. e^ix is a rotation by x. e^i4x is a rotation by 4x.
Have you sketched what you need to do?
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1 week ago
#10
(Original post by mqb2766)
For the first point, you've just done demoivre. e^ix is a rotation by x. e^i4x is a rotation by 4x.
Have you sketched what you need to do?
ah right. yeah i've sketched it.
So i need to sum e^0 then e^(2pi/9) then e^(4pi/9) then e^(6pi/9)...and that's it?

(Original post by Itsmikeysfault)
Using complex numbers just makes the problem more complex that it needs to. You could get the answer by drawing a diagram and using trig
this was from the complex numbers part of the textbook so im just trying to solve it using CN.
Last edited by Htx_x346; 1 week ago
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1 week ago
#11
(Original post by Htx_x346)
ah right. yeah i've sketched it.
So i need to sum e^0 then e^(2pi/9) then e^(4pi/9) then e^(6pi/9)...and that's it?

this was from the complex numbers part of the textbook so im just trying to solve it using CN.
They want you to solve the problem by using complex numbers to represent translations and rotations. That sounds about right and you should note it is a simple geometric series so you can use the standard result.
Last edited by mqb2766; 1 week ago
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1 week ago
#12
(Original post by mqb2766)
They want you to solve the problem by using complex numbers to represent displacements and rotations. That sounds about right and you should note it is a simple geometric sequence.
thanks.
But would it not be slightly more tedious using the sum of geometric sequences equation in this case when you can just add them?
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1 week ago
#13
(Original post by Htx_x346)
thanks.
But would it not be slightly more tedious using the sum of geometric sequences equation in this case when you can just add them?
Id go for the series, but why not try both and see what you find out.
Its not going to be that long either way.

Last edited by mqb2766; 1 week ago
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1 week ago
#14
anyone know where the OP got the question from
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1 week ago
#15
anyone know where the OP got the question from
It's a question from the edexcel further maths core pure 2 textbook
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1 week ago
#16
(Original post by mqb2766)
Id go for the series, but why not try both and see what you find out.
Its not going to be that long either way.

thanks

also, just to clarify, what exactly was the reason for (e^2pi/9)^4 being incorrect? thanks again
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1 week ago
#17
(Original post by Htx_x346)
thanks

also, just to clarify, what exactly was the reason for (e^2pi/9)^4 being incorrect? thanks again
Its e^(i8pi/9). So a distance of 1 from the origin.
Last edited by mqb2766; 1 week ago
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1 week ago
#18
(Original post by mqb2766)
Its e^(i8pi/9). So a distance of 1 from the origin.
thank you have a good evening
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1 week ago
#19
(Original post by Itsmikeysfault)
Using complex numbers just makes the problem more complex that it needs to. You could get the answer by drawing a diagram and using trig
Can't see why using complex numbers makes it more complex, though obviously you can do it with straight geometry/trig. With complex numbers, after writing down the geometric series expression, you can "centre" both the numerator and denominator and pretty much write down the answer?
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1 week ago
#20
(Original post by mqb2766)
Can't see why using complex numbers makes it more complex, though obviously you can do it with straight geometry/trig. With complex numbers, after writing down the geometric series expression, you can "centre" both the numerator and denominator and pretty much write down the answer?
I solved it both ways. Using trig was faster and required less thinking. Most further maths students would agree
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