Urgent Help!!! (1 Viewer)

[red-butterfly]

can someone show me the working for the following questions...?

1. On an argand diagram the points A and B represent the complex numbers z1 = i and z2 = 1/√2(1 + i). Show that arg(z1 + z2) = 3π/8

2. Use the vector representation of z1 and z2 on an Argand diagram to show that
a) If |z1| = |z2| ,then (z1 + z2)/(z1 - z2) is imaginary

b) If 0 < arg z2 < arg z1 < π/2, andarg (z1 - z2) - arg(z1 - z2) = π/2


THANKS ^^

 

[vds700]

can someone show me the working for the following questions...?

1. On an argand diagram the points A and B represent the complex numbers z1 = i and z2 = 1/√2(1 + i). Show that arg(zi + z2) = 3π/8

2. Use the vector representation of z1 and z2 on an Argand diagram to show that
a) If |z1| = |z2| ,then (z1 + z2)/(z1 - z2) is imaginary

b) If 0 < arg z2 < arg z1 < π/2, andarg (z1 - z2) - arg(z1 - z2) = π/2


THANKS ^^

do u mean arg(z1 + z2) = 3π/8?

 

[tacogym27101990]

can someone show me the working for the following questions...?

1. On an argand diagram the points A and B represent the complex numbers z1 = i and z2 = 1/√2(1 + i). Show that arg(zi + z2) = 3π/8

2. Use the vector representation of z1 and z2 on an Argand diagram to show that
a) If |z1| = |z2| ,then (z1 + z2)/(z1 - z2) is imaginary

b) If 0 < arg z2 < arg z1 < π/2, andarg (z1 - z2) - arg(z1 - z2) = π/2


THANKS ^^

bit rusty on argand diagrams but ill give it a shot

1. arg(z1+z2)= arg(i+(1/rt2)+i(1/rt2))
=arg ((1/rt2) + i(1+rt2/rt))
= tan^-1(1+rt2)
=3pi/8

2.a) let O be origin, A be z1, B be z1+z2, Cbe z2
if |z1|=|z2| then OABC is a rhombus
therefore the diagonals intersect at right angles
so z1+z2= ki(z1-z2)
so (z1+z2)/(z1-z2) = ki(z1-z2)/(z1-z2) = ki which is imaginary

b) whats the question?

somebodytell me if im wrong cause i havent done a question like this since last year

 

[red-butterfly]

yeh = = z1!!

 

[u-borat]

wats z2 equal to?

are the brackets on the top or the bottom?

 

[red-butterfly]

ekk let me reword 2b) If 0 < arg z2 < arg z1 < π/2, andarg (z1 - z2) - arg(z1 - z2) = π/2, prove that |z1| = |z2|

 

[tacogym27101990]

ekk let me reword 2b) If 0 < arg z2 < arg z1 < π/2, andarg (z1 - z2) - arg(z1 - z2) = π/2, prove that |z1| = |z2|

ok so that means the diagonals of the parallelogram OABC intersect at right angles, meaning the that the parallelogram is a rhombus
and a rhombus has equal sides
therefore |z1|=|z2|

EDIT: wait is one of those supposed to be arg(z1+z2)?
cause thats what i did

so if its not totally disregard this working

 

[tacogym27101990]

so have you got the answers red butterfly??

 

[red-butterfly]

i dunno = = stupid handout... i'll just ask my teacher on monday ... but thanks for your help ^ ^

 

[red-butterfly]

yehh

 

[red-butterfly]

ok so that means the diagonals of the parallelogram OABC intersect at right angles, meaning the that the parallelogram is a rhombus
and a rhombus has equal sides
therefore |z1|=|z2|

EDIT: wait is one of those supposed to be arg(z1+z2)?
cause thats what i did

so if its not totally disregard this working

yeh its suppose to be arg(z1 + z2)

 

[tacogym27101990]

o wow are you starting 4 unit maths already??
we didnt start till like 5 weeks into 4th term last year

 

[red-butterfly]

really?? well we started about 3 weeks ago. its more like the trial period so we have enough time to drop it if we dont like it.