Complex number questions (1 Viewer)

[Marc26]

I got stuck on these:

Question 2:
a) The origin O and the points A, B, C representing complex numbers z, 1/z, z+(1/z) respectively are joined to form a quadrilateral. Write down the condition or conditions for z so that OABC will be:
i) a rhombus
ii) a square

b) A, B, C are the angles of a triangle. Show that:
(CosA + i SinA)(CosB + i SinB) - (CosC + i SinC) = 0

 

[pokka]

For Question 2 a), think about the properties of a rhombus and a square. A rhombus has equal sides and diagonals are perpendicular to each other and a square has these properties and also their angles are 90 degrees. Now think about how you can relate that for conditions in terms of z. Perhaps drawing a diagram of your quadrilateral OABC (with labelled sides in terms of z) can help you visualise it and figure it out.

For Question 2 b), expand the expression on the left and group all the real and imaginary parts (but do this for only (CosA + i SinA)(CosB + i SinB)). For that part, see if you can simplify it (hint: it should simplify for something containing (A+B) ). After you do that, think about how (A+B) and C are related in terms of angles in a triangle and then it should get you there

 

[Sy123]

I got stuck on these:

Question 2:
a) The origin O and the points A, B, C representing complex numbers z, 1/z, z+(1/z) respectively are joined to form a quadrilateral. Write down the condition or conditions for z so that OABC will be:
i) a rhombus
ii) a square

b) A, B, C are the angles of a triangle. Show that:
(CosA + i SinA)(CosB + i SinB) - (CosC + i SinC) = 0

Ok look at it this way, for a:

What is 1/z?

When we look at

By de Moivre's theorem

Now when we add these two complex numbers, we have to do so using vector division to get z+1/z
This is done by constructing a parallelogram off the two complex numbers, parallel to the opposite vector.

Now if it were a rhombus, the diagonals must bisect the angles of a rhombus, hence it must bisect the angle BOA (C being z+1/z)
But hang on a second, the real axis bisects this angle (into two units of theta (by the arguments of z and 1/z))

Hence for it to be a rhombus, z+1/z MUST be real

Hence Im(z+1/z)=0

Hence Im(z)+Im(1/z)+0

let z=x+iy, you get




Hence our modulus of z must be 1. Now you must be thinking, why didnt I just do this

(adjacent sides of a rhombus must be equal). However in the end it is all the same, the restriction is that mod z =1

for b, the angle must be a right angle, hence

Now

To sum it up: mod z = 1 arg z=pi/4



For b:

Just do some technical expansion:




But A+B=180-C (angle sum of triangle)





Question wrong perhaps?

 

[HeroicPandas]

nvm