Couple of Complex Quickies (1 Viewer)

[yrtherenonames]

So we've got trials in 5 odd weeks and its time to start up the old complex numbers routine. Here's two that I can't get
1.a)ii)Let z=pi/5
(Hence) Show that (1+sinz+icosz)^5+i(1+sinz-icosz)^5=0

In part (i) we were asked to show that (1+sinA+icosA)/(1+sinA-icosA)=sinA+icosA

I used part (i) a bit but I'm not really sure where to go.

2a) Sketch [z-3]=3 (no problem)

b) Show that arg(z-3)=arg(z^2) (problem)
I suspect I should be able to do that but again I'm not sure where to go.

 

[Ogden_Nash]

1)ii)

RTP:
(1+sinz+icosz)^5+i(1+sinz-icosz)^5=0
(1+sinz+icosz)^5= -i(1+sinz-icosz)^5
(1+sinz+icosz)^5/(1+sinz-icosz)^5 = -i
[(1+sinz+icosz)/(1+sinz-icosz)]^5 = -i

LHS = (sinz+icosz)^5 {from i}
.......= sin(pi) + icos(pi)
.......= -i
.......= RHS


2b)

The locus in a) is a circle centre (3,0), radius 3 and you are required to prove:
....arg(z-3) = arg(z²)
ie arg(z-3) = 2arg(z)

Now, in your original locus, construct a radius from centre C(3,0) to any point on the circumference (P say). Now, angle PCx = 2*angle POx (angle at centre is twice angle at circumference where x is the positive x axis).
Angle PCx represents arg(z-3) and angle POx represents arg(z), therefore:
arg(z-3) = 2arg(z)

 

[who_loves_maths]

Originally Posted by yrtherenonames
1.a)ii)Let z=pi/5
(Hence) Show that (1+sinz+icosz)^5+i(1+sinz-icosz)^5=0

In part (i) we were asked to show that (1+sinA+icosA)/(1+sinA-icosA)=sinA+icosA

I used part (i) a bit but I'm not really sure where to go.

hi yrtherenonames,

for your first question:

(1+sinz+icosz)^5+i(1+sinz-icosz)^5=0 ---> divide throughout by (1+sinz-icosz)^5
---> [(1+sinz+icosz)^5]/[(1+sinz-icosz)^5] + i = 0 = [(1+sinz+icosz)/(1+sinz-icosz)]^5 + i
now use part (i) to get ---> 0 = (sinz + icosz)^5 + i ;
but note that (sinz + icosz) = i(cosz - isinz) = i(cos(-z) + sin(-z))
hence, 0 = [i(cos(-z) + sin(-z))]^5 + i ;
and by DMT, (cos(-z) + sin(-z))^5 = cos(-5z) + sin(-5z),
therefore, 0 = i^5(cos(-5z) + sin(-5z)) + i = icis(-5z) + i ---> 0 = cis(-5z) + 1
and from that you get z = pi/5 as a root of that.

and so for the original (1+sinz+icosz)^5+i(1+sinz-icosz)^5=0, it is true when z=pi/5 since that is a root.

2b) Show that arg(z-3)=arg(z^2) (problem)
I suspect I should be able to do that but again I'm not sure where to go.

for your question 2:

do this geometrically.
the locus of [z-3] is a circle of radius 3 with centre C(3, 0), and it just touches O(0, 0).
now, let a point on this circle be 'Z', hence you can construct the triangle OCZ ; and let arg(z) = @
now, arg(z) = @ = < ZOC , and arg(z-3) = < ZOS, where 'S' is just any point on the positive x-axis far away from 'C' in the direction O->C.
now, triangle OCZ is isosceles with OC = ZC (radii of circle), hence, < ZOC = @ = < OZC (equal angles of isos. triangle); hence, < ZOS = < ZOC + < OZC = 2@ (exterior angle is sum of two oppo. interior angles).

and so < ZOS = arg(z-3) = 2@ = 2(< ZOC) = 2arg(z) = arg(z^2) ;
ie. arg(z-3) = arg(z^2)


hope this helps

 

[yrtherenonames]

Thanks guys