more complex numbers ^^ (1 Viewer)

[noobking]

 

[haque]

for the first one, when using the parallelogram method of adding vectors u get a rhombus as the shape with modulus z1=modulus z2, the vectors z1-z2 and z1+z2 are the diagonals-the diagonals of a rhombus are always at right angles thus the ratio in the question is in the form +- ki i.e purely imaginary.

 

[haque]

For the second part, z1- (z1-2iz2)/(1-2i)=(1-2i)z1 -z1 +2iz2/(1-2i)
=-2iz1+2iz2/(1-2i)=2i(z2-z1)/1-2i and this vector is perpendicular to the vector z1-2z2/1-2i -z2 i.e one side of the triangle formed by these three points makes right angles with another side, hence it is a right angled triangle(i wouldn't advise u to do it as briefly as i did-make sure u give full reasons and show as much working as possible esp a diagram)

For the third one a diagram is a must and when u draw it, the origin, the centre of the circle and A form an isosceles triangle with the sides CO(C is centre)and CA being each 2 units. then if r=modulus z then use the sine rule: r/sin(180-2@) =2/sin@
r/2sin@cos@=2/sin@ i.e r=4cos@
Editamn i'm hopeless with computers, couldn't even put on a spoiler,
modulus 1/z=1/4cos@ and arg(1/z)=-@
and 1/z=1/4cos@(cis(-@)=1/4cos@(cos@-isin@)=1/4 -isin@/4cos@ i.e the real part of the point P is constant so P lies on a straight line(if 1/z=x+iy then x=1/4 and is constant so then that's the equation of the line).
Another way we could have done it was let z=x+iy. We know that x^2+y^2-4x+4=4 i.e x^2+y^2=4x then 1/z=1/x+iy = x-iy/(x^2+y^2) if 1/z=u+iv then u=x/(x^2+y^2)=x/4x=1/4 i.e u constant as v varies.

Edit: there was a typo

 

[bos1234]

haque do u have ms word?

 

[haque]

Lol yeah(i know i should have done it there then pasted it but i couldn't be bothered)

 

[jyu]

:wave: