Thread: Complex D Name F D Number

1. Complex D Name F D Number

QUESTION 1)
(1+i)(1+2i)(1+3i)........(1+ni)=A+iB
show dat : tan^-1(1)+tan^-1(1/2)+tan^-1(1/3)..........tan^-1(1/n)=tan^-1(B/A)
QUESTION 2)
if x=Cos a +i Sin a;y=Cos b+i Sin b;z=Cos c +i Sin c,prove dat
(1)1/x+1/y+1/z=0 whenevr x+y+z=0
(2)xy+yz+zx=0 weneva x+y+z=0
(3)xy/z+z/xy=2cos(a+b+c)
ok basicaly al i noe is u shud use complex numbr properties..
so plzzzz help

2. Originally Posted by mathsjerk101
fine here is d question:
QUESTION 1)
(1+i)(1+2i)(1+3i)........(1+ni)=A+iB
When you have,
$1+ai$
You can write,
$\sqrt{1+a^2} \left( \frac{1}{\sqrt{1+a^2}}+\frac{a}{\sqrt{1+a^2}} i \right)$
Thus,
$sqrt{1+a^2} (\cos \theta_a + i \sin \theta_a )$
Where,
$\theta_a = \tan^{-1} a$

Thus, we have,
$\sqrt{1+1^2}(\cos \theta_1 +i\sin \theta_1)\cdot ... \cdot \sqrt{1+n^2}(\cos \theta_n+i\sin \theta_n )$
Thus,
$\sqrt{(1+1^2)\cdot ... \cdot (1+n^2)} (\cos (\theta_1 +... + \theta_n )+i\sin (\theta_1 + ... + \theta_n) )$

Thus,
$A=\sqrt{(1+1^2)...(1+n^2)}\cdot \cos (\tan^{-1} 1+...+\tan^{-1} n)$

$B=\sqrt{(1+1^2)....(1+n^2)}\cdot \sin (\tan^{-1} 1+...+\tan^{-1} n)$

3. Originally Posted by mathsjerk101
QUESTION 2)
if x=CiS a,y=CiS b,z=CiS c,prove dat
(1)1/x+1/y+1/z=0 whenevr x+y+z=0
(2)xy+yz+zx=0 weneva x+y+z=0
(3)xy/z+z/xy=2cos(a+b+c)
I do not understand the problem.

Look at #2)
x=1,y=2,z=-3
It does not work.

4. Originally Posted by ThePerfectHacker
When you have,
$1+ai$
You can write,
$\sqrt{1+a^2} \left( \frac{1}{\sqrt{1+a^2}}+\frac{a}{\sqrt{1+a^2}} i \right)$
Thus,
$sqrt{1+a^2} (\cos \theta_a + i \sin \theta_a )$
Where,
$\theta_a = \tan^{-1} a$

Thus, we have,
$\sqrt{1+1^2}(\cos \theta_1 +i\sin \theta_1)\cdot ... \cdot \sqrt{1+n^2}(\cos \theta_n+i\sin \theta_n )$
Thus,
$\sqrt{(1+1^2)\cdot ... \cdot (1+n^2)} (\cos (\theta_1 +... + \theta_n )+i\sin (\theta_1 + ... + \theta_n) )$

Thus,
$A=\sqrt{(1+1^2)...(1+n^2)}\cdot \cos (\tan^{-1} 1+...+\tan^{-1} n)$

$B=\sqrt{(1+1^2)....(1+n^2)}\cdot \sin (\tan^{-1} 1+...+\tan^{-1} n)$
You do realise that this was not the question don't you?

The question was to show that:

atan(1)+atan(1/2)+atan(1/3)+...+atan(1/n)=atan(B/A)

RonL