# complex no. Q (1 Viewer)

#### Hikari Clover

##### Member
the points A, B ,C represent the complex number z, 1/z , z+1/z respectively.
O is the origin. the quadrilateral OACB is formed by joining the points.
write down the condition(s) for z so that OACB is a
1)rhombus
2)square

|z-i| - |z+i| = 4
is it the locus of a hyperbola ?
in what situation it will become a branch of a hyperbola :wave:

#### lilfrenchi

##### New Member
i think for 2) |z|=1
im not really sure tho

#### TesseracT22

##### New Member
i think 1) is arg(z) =/= kpi/2 where k is any integer.

2) |z| = 1 and also arg(z) =/= kPi/2

#### TesseracT22

##### New Member
for the second bit, |z-i| - |z+i| = 4 satisifes, |PS - PS'| = 2a.
So it's a hyperbola with foci at i and -i.

It would become a branch only, if Im(z)>0 or if Im(z)<0

#### bos1234

##### Member
let z = cos@+isin@

1/z = z^-1 = cos@ - isin@ = z (conjugate)

so lot a point z any where on the argand diagram and plot the onjugate

z^1+z^-1 = 2cos@ which lies on the x-axis

u know the rule? z^n+z^(-n) = 2ncos@

Now apply rhombus properties.
If OABC is a rhombus then vector OA = vector OB
i.e. |z|=1/|z|

I think thats how you do it

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for square? whats the difference between qsuar =e and rhombus?

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for your last qn look at the 2006 paper for complex numbers(last qn on ellipse) and look at the solution. they have a good solution to it and u can use it for this

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#### Hikari Clover

##### Member

guys,
|z-i| - |z+i| = a
is just a branch of a hyperbola ...........

#### Tmer

##### New Member
for a)
z was not specified a mod of 1 so if you are calling it cos@+isin@ then you are changing the question..

1/z is = z(conjugate)/|z| (multiply top n bottom by z conjugate)

this means 1/z has the negative arg(z) and a modulus less then or more then z unless |z|=1

since arg(z) = -arg(1/z) then arg(z+1/z) = 0 it always lies along the x axis..

now, the rule for a rhombus is all sides equal, thinking in vectors this mean that z and 1/z have an equal modulus one condition, the second condition is arg(z) is less then 90 degrees..
b) for OACB to be a square all interior angles must be 90 so
change the condition arg(z)=45 or better yet remember that a multiplication by i is a counter clockwise rotation by 90 degrees so i x 1/z = z ---> z^2=i

as for last part i am not sure how to draw it even? id be glad if some 1 explained it to me please...

#### kony

##### Member
it is obviously only one branch because the distance to one focus is being subtracted from the other.

which means if it was |z-i| - |z+i| = +-4 would indicate both branches

#### Tmer

##### New Member
can some 1 explain why it is a hyperbola? i can see why |z+x+i|+|z-x-i|=a is an ellipse in the complex field, but i cant exactly see why this is a hyperbola?

#### Hikari Clover

##### Member

|z+x+i|+|z-x-i|=a
ellipse

|z+x+i|- |z-x-i|=a

hyperbola,actually 1 branch only.....