FP2 Complex Numbers
Please could someone explain how this dotted line represents the complex number given?
Thanks for any help
Thanks for any help
Original post by PhyM23
Please could someone explain how this dotted line represents the complex number given?
Thanks for any help
Thanks for any help
The complex number is the line joining to . To see this, sketch out two points and pretend they're vectors. So , kinda like that.
In this case, you have and
Original post by Zacken
The complex number is the line joining to . To see this, sketch out two points and pretend they're vectors. So , kinda like that.
In this case, you have and
In this case, you have and
Ah okay, but how do you know that lies on the circle?
Original post by PhyM23
Ah okay, but how do you know that lies on the circle?
Like it said, because it's length is equal to the radius of the circle.
Original post by Zacken
Like it said, because it's length is equal to the radius of the circle.
That's the bit I don't understand. It doesn't say that in the question itself. I'm not sure what reasoning there is to say that this is true.
Original post by PhyM23
That's the bit I don't understand. It doesn't say that in the question itself. I'm not sure what reasoning there is to say that this is true.
Ah, well that changes thins entirely! Basically - you have a semicircle. The radius of the semi-circle starting from the marked position is the perpendicular bisector of and which is
Original post by Zacken
Ah, well that changes thins entirely! Basically - you have a semicircle. The radius of the semi-circle starting from the marked position is the perpendicular bisector of and which is
I get this more now, but why can't be a point that's not on the circle? I.e how do you know it lies on the edge of the circle?
Original post by PhyM23
I get this more now, but why can't be a point that's not on the circle? I.e how do you know it lies on the edge of the circle?
Because that's what the start of the question defines it as. lies on .
Original post by Zacken
Because that's what the start of the question defines it as. lies on .
Of course. Can't believe I missed that! Thank you!
Original post by PhyM23
Of course. Can't believe I missed that! Thank you!
No problem!
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