Complex Number question (1 Viewer)

[Cobra]

Hey guys, im stumped on how to do this so if you could help me, it would be much appreciated:

On an Argand Diagram, the Point A represents the real number 1. O is the Origin, and the point P represents the complex number Z, which satisfies the condition:

Arg (Z - 1) = 2 Arg Z

a) Show this information on a diagram, and deduce that the triangle OPA is isosceles.

b) Deduce that the locus of P is a circle, and show this on your diagram.

I also cant do this question

1) Let ß be a cube root of -1, where ß is not real. Show that ß^2 = ß - 1

 

[Mountain.Dew]

Hey guys, im stumped on how to do this so if you could help me, it would be much appreciated:

On an Argand Diagram, the Point A represents the real number 1. O is the Origin, and the point P represents the complex number Z, which satisfies the condition:

Arg (Z - 1) = 2 Arg Z

a) Show this information on a diagram, and deduce that the triangle OPA is isosceles.

b) Deduce that the locus of P is a circle, and show this on your diagram.

I also cant do this question

1) Let ß be a cube root of -1, where ß is not real. Show that ß^2 = ß - 1

the complex number one i can do, but i just cant seem to express it correctly without a diagram.

for the roots one ==>

if B is a cube root of -1, then B^3 = -1 ==> B^3 + 1 = 0, therefore (B+1)(B^2 - B + 1)=0

since B is a complex number, B + 1 cannot = 0. therefore, B^2 - B + 1=0

so ß^2 = ß - 1

 

[Cobra]

cheers mountain dew, that was a great help, it'll give me an extra 2 marks
Thanx dude =D

 

[Riviet]

From my diagram, let θ be the argument of Z (angle AOP). Since arg(z-1) refers to the angle that z makes with P(1,0) and is double the angle that z makes with the Re(z) axis, arg(z) = θ. Now the exterior angle of a triangle is equal to the sum of the two opposite interior angles, implying that angle OAP = arg(z-1) - angle AOP = 2θ - θ = θ.
Therefore, triangle OAP is isosceles.
Now for part b), i'm not sure how to go about it algebraically, but look at the diagram and imagine a circle that passes through O and A, with centre P, where OP and OA are two of the radii of the circle. Hope that helps.

 

[Mountain.Dew]

From my diagram, let θ be the argument of Z (angle AOP). Since arg(z-1) refers to the angle that z makes with P(1,0) and is double the angle that z makes with the Re(z) axis, arg(z) = θ. Now the exterior angle of a triangle is equal to the sum of the two opposite interior angles, implying that angle OAP = arg(z-1) - angle AOP = 2θ - θ = θ.
Therefore, triangle OAP is isosceles.
Now for part b), i'm not sure how to go about it algebraically, but look at the diagram and imagine a circle that passes through O and A, with centre P, where OP and OA are two of the radii of the circle. Hope that helps.

to get it algebraically, just do z=x+iy

arg(z-1) = 2arg z

arg(x-1+iy) = 2(arg)(x+iy)

arg(x-1+iy) = arg(x+iy)^2

arg(x-1+iy) = arg(x^2 - y^2 + 2xyi)

so, skipping the tan inverse part... y/(x-1) = 2xy/ (x^2 - y^2), we have y=0

x^2 - y^2 = 2x^2 - 2x

x^2 - 2x + 1 + y^2 = 1

(x-1)^2 + y^2 = 1

TADA!

 

[Riviet]

Ah so that's how you do it, the inverse tan bit cancelling out certainly makes sense now. =D

 

[Mountain.Dew]

Ah so that's how you do it, the inverse tan bit cancelling out certainly makes sense now. =D

cheers mountain dew, that was a great help, it'll give me an extra 2 marks
Thanx dude =D

=D glad i helped. *bows to both ppl*

 

[Mill]

The first question is actually quite interesting in that there is several ways that you can think about it.

It's also connected to the circle geometry theorem that the angle at the centre of a circle is twice the angle at the circumference.

 

[Cobra]

cheers fellaz. I actually worked it out another way, using trig and working out question b before a thanx again though, you guys were very helpful.