another complex number question (2 Viewers)

[pLuvia]

From cambridge

Ex2.3
2(a) Use the vector representation of |z₁| = |z₂|, then (z₁ + z₂) / (z₁ - z₂) is imaginary.
(b) If 0 < argz₂ < argz₁ < pi/2, and arg(z₁ - z₂) - arg(z₁ + z₂) = pi/2, then |z₁| = |z₂|.

4. On an argand diagram the points P and Q represent z and z+iz respectively. Show that OPQ is a right angled triangle.

 

[richz]

2a) draw the |z₁|and |z₂|vectors out, and wat do u get? u get an isos triangle rite? then u draw out z₁+z₂ and z₁-z₂. Wat angle does it create? this angle is on the imag axis.

b) using part one u should see its a rhombus, so wats a property of a rhombus?

4. draw out the vectors, wats iz? isnt it z moved 90 degrees anticlockwise so wouldnt it be a right angled triangle? with right angle at POQ..

 

[pLuvia]

Oh yeh, I understand 4 because multiplying by i rotates the vector 90 degrees anti clockwise

2a It's not necessarily on the imag. axis

By the way can you use mathematical induction to prove that one thing equals the other?

Like prove A = B or is it only for proving that it satisfies a certain condition?

 

[richz]

Kadlil, why wud u use induction, it would take a while.

anyway back to q 2, ill give u more info.

arg(z₁ + z₂) / (z₁ - z₂) = arg(z₁ + z₂) - arg (z₁ - z₂).

EDIT: since |z₁| = |z₂| therefore OABD (D is z₁+z₂) is a rhombus or square, so wouldnt the diagonals intersect at right angles

try that

 

[pLuvia]

Is it because i = 90 degrees then the complex number is imaginary?

 

[pLuvia]

Show that ||z1| - |z2|| <[u/] |z1 + z2|. State the condition for equality to hold.

 

[pLuvia]

Use De Moivre's theorem to solve the equation z⁵ = 1. Show that the points representing the five roots of this equation on an Argand diagram form the vertices of a regular pentagon of area 5/2 sin 2pi/5 and perimeter 10sin pi/5

 

[richz]

Use De Moivre's theorem to solve the equation z⁵ = 1. Show that the points representing the five roots of this equation on an Argand diagram form the vertices of a regular pentagon of area 5/2 sin 2pi/5 and perimeter 10sin pi/5

if z=rcis@
.'. z⁵=r⁵cis5@ (De Moivres)
.'. r⁵cis5@=1
r⁵=1
.'.r=1 (mod), so all mods the same, so shud form a regular polygon
5@=0+2kpi (k any integer, ie. in this case -1,1,2,-2 and 0)
@=(2kpi)/5

Draw out the vectors and they shud form a regular pentagon. Then find area of each isos triangle using sine rule and multiply by 5.

hope that helps

 

[pLuvia]

thanks again