Ok so I have 2 points

 

A = 2 + 3j

C = 4 + 7j

 

on our argand diagram they make an equailateral triangle with a point (or more precisely either of 2 points) which I'll call z

 

Now I know;

[math]

 

ABC = arg ( \lambda ) = arg (\frac{a - b}{c-b})

 

 

 

[/math]

so

[math]

arg(\lambda) = \frac{\pi}{3}

 

[/math]

 

and

 

[math]

\lambda = \frac{a-b}{c-b}

[/math]

[math]

\lambda = \frac{(z-(4+7j))-((4+7j)-(2+3j))}{(z-(2+3j))-((4+7j)-(2+3j))}

[/math]

[math]

\lambda = \frac{z-6-11j}{z-4-j}

[/math]

 

 

No where do I go from here; solve simultaneously, or am I using the completely wrong method?

 

I've attatched a sketch of the argand diagram.

Ok so I have 2 points

 

A = 2 + 3j

C = 4 + 7j

 

on our argand diagram they make an equailateral triangle with a point (or more precisely either of 2 points) which I'll call z

 

Now I know;

[math]

 

ABC = arg ( \lambda ) = arg (\frac{a - b}{c-b})[/math]

I'm not familiar with this. "ABC" is a real number. What number? The cosine of the angle between sides?

 

I would be more inclined to use modulus than arg: |A- C|= |-2- 4j|= [math]\sqrt{4+ 16}= \sqrt{20}[/math]. Since this is an equilateral triangle, we must have also |z- A|= |z- C|=[math]\sqrt{20}[/math] also. Letting z= x+ yj, [math]\sqrt{(x-2)^2+ (y-3)^2}= \sqrt{20}[/math] and [math]\sqrt{(x-4)^2+ (y-7)^2}= \sqrt{20}[/math].

 

Of course, those immediately give [math](x- 2)^2+ (y-3)^2= 20[/math] and [math](x- 4)^2+ (y- 7)^2= 20[/math]. You should be able to see that, after you multiply those out, the [math]x^2[/math] and [math]y^2[/math] terms cancel and you have two linear equations to solve for x and y.

 

 

so

[math]

arg(\lambda) = \frac{\pi}{3}

 

[/math]

 

and

 

[math]

\lambda = \frac{a-b}{c-b}

[/math]

[math]

\lambda = \frac{(z-(4+7j))-((4+7j)-(2+3j))}{(z-(2+3j))-((4+7j)-(2+3j))}

[/math]

[math]

\lambda = \frac{z-6-11j}{z-4-j}

[/math]

 

 

No where do I go from here; solve simultaneously, or am I using the completely wrong method?

 

I've attatched a sketch of the argand diagram.

if 2a+7x(95z^2)=67

 

how do u find a?

By asking for z and x.

no, like in the form a=?z +/- ?x

1) substract the 2nd addend on the left-hand side from both sides

2) divide both sides by 2