How to solve this? Dividing complex numbers via trigonometric means?

Question

Okay, we have (1- i)/(√3 - i).

I did exactly like I'm supposed to and got:

√2 [ cos (√2/2) + i sin ( -√2/2) ]
------------- ------------- --------------
2 [ cos (√3/2) + i sin ( -1/2) ]

=

√2 [ cos (7pi/4) + i sin (7pi/4) ]
------------- ------------- --------------
2 [ cos (11pi/6) + i sin (11pi/6) ]

And now I have to divide the √2 by 2 and subtract in the following manner:

[ cos (7pi/4 - 11pi/6) + i sin (7pi/4 - 11pi/6) ]

and I get:

√2/2 [ cos ( - pi/12) + i sin ( - pi/12) ]

Now before I go and do out the sum/difference of two cos and sin, etc. because I can't find a trig table with -pi/12, can someone just tell me if my answer is right so far?

Thank you so much! I'll pick a best answer TODAY!!!!!!!

Where the **** did you get a 4 from?

Answer 1

z = (1 - i) / [sqrt(3) - i]

Numerator = 1 + (-1)i
x = 1 , y = -1 , r = sqrt(2)
cos(t) = x / r = 1 / sqrt(2) , sin(t) = y / r = -1 / sqrt(2)
Therefore, t = -45º = 315º = 7pi/4 (angle is in 4th quadrant)
Thus, numerator = sqrt(2)[cos(7pi/4) + i*sin(7pi/4)]

Denominator = sqrt(3) + (-1)i
x = sqrt(3) , y = -1 , r = 2
cos(t) = x / r = sqrt(3) / 2 , sin(t) = y / r = -1 / 2
Therefore, t = -30º = 330º = 11pi/6 (angle also in 4th quadrant)
Thus, denominator = 2[cos(11pi/6) + i*sin(11pi/6)]

z = sqrt(2)[cos(7pi/4) + i*sin(7pi/4)] /
{2[cos(11pi/6) + i*sin(11pi/6)]}

On dividing, we get :

z = [sqrt(2) / 2][cos(7pi/4 - 11pi/6) + i*sin(7pi/4 - 11pi/6)]

z = [sqrt(2) / 2][cos(-pi/12) + i*sin(-pi/12)]

I get the same answer as you do. Confirmed.

Answer 2

let z =(1- i)/(√3 - i).

let's simplify this

(1-i) ( √3 +i) /{ 4)
=( √3 - i√3 +i + 1)/4 = (1 + √3)/4 + i ( 1 -√3 )/4 = x + iy
x = (1 + √3)/4
y = ( 1 -√3 )/4

r^2 = {( 1 + 3 + 2√3) + ( 1 +3 - 2√3) }/16 =
1/2

r = √2 / 2
z = √2 / 2 { (1 + √3)/4 )√2 / 2 + i (√2 / 2 )( 1 -√3 )/4 } =

√2 / 2 ( let z =(1- i)/(√3 - i).

let's simplify this

(1-i) ( √3 +i) /{ 4)
=( √3 - i√3 +i + 1)/4 = (1 + √3)/4 + i ( 1 -√3 )/4 = x + iy
x = (1 + √3)/4
y = ( 1 -√3 )/4

r^2 = {( 1 + 3 + 2√3) + ( 1 +3 - 2√3) }/16 =
1/2

r = √2 / 2
z = √2 / 2 { (1 + √3)/4 )√2 / 2 + i (√2 / 2 )( 1 -√3 )/4 } =

√2 / 2 ( (√2 / 8) + √6/8 + i ( (√2 / 8) - √6/8) } =

(√2 / 4 ){ (√2 + √6)/4 - i (√6 - √2)/4 }

Note that sin (5Pi/12) = (√2 + √6)/4 and
cos (5pi/12) = (√6 - √2)/4

z = (√2 / 4 ) ( sin (5pi/12) -i cos(5pi/12)}

Answer 3

Why use trigonometric means?
(1 - i) / (√3 - i)
(1 - i)(√3 + i) / (√3 - i)(√3 + i)
√3 + i - √3i - i²) / (3 - i²)
(√3 + 1 + i - √3 i ) / 4
( √3 + 1 + (1 - √3) i ) / 4