complex number question with tan (1 Viewer)

[lsdpoon1337]

If z= cos a + isin a, prove:

1/(z+1) = (1/2)(1 - itan a/2 )

thanks

 

[Js^-1]

If z= cos a + isin a, prove:

1/(z+1) = (1/2)(1 - itan a/2 )

thanks

z = cos a + isin a

LHS = 1/(z+1)
= 1/([1+cos a] + isin a)
=[(1 + cos a) - sin a)] / [(1+cos a) - isin a][(1+cos a) + isin a]
...Can't be bothered working out the rest but I think thats how you start...

 

[Yamiyo]

First rationalise the denominator by multiplying top and bottom by (cosa+1)-isina

Then you get
((cosa+1)-sina)/((cosa+1)^2)+(sina)^2)

Tidy that up a bit and you get
(cosa+1-isina)/(2+2cosa)

Take out the two from the denominator
(1/2)(cosa+1-isina)/(cosa+1)
=(1/2)(1-(isina/(cosa+1))

Now use the t results to tidy that up and get the required result.

 

[lsdpoon1337]

yes, I got up to that part, but can't seem to solve the question from there

 

[lsdpoon1337]

my bad, lol

thanks

 

[zinc]

replace sina with 2sin(a/2)cos(a/2) and cosa with 2cos(squared)(a/2)-1 (using half angle formulas) then it cancels down. Sorry if that doesn't make much sense (i dont know how to write maths on the computer)

 

[vds700]

you dont really need to rationallise the denominator, heres my solution,

1/(1+z)
=1/(1 + cosa + isina)
=1/(1 + 2cos^2 (a/2) -1 +2isin(a/2)cos(a/2))
=1/(2cos^2 (a/2) +2isin(a/2)cos(a/2))
=1/[2cos(a/2)(cos(a/2) + isin(a/2)]
=1/2cos(a/2) . cis(a/2)^-1
=1/2cos(a/2) . (cos(a/2) - isin(a/2) )using De Moivre's theorem
=1/2 . (1 - itan(a/2)

 

[Js^-1]

That's a nice solution...
Looks nice and quick.

 

[untouchablecuz]

LHS = 1/(z+1)

= 1/(cos a + isin a)

Using t (t = tan a/2) formula on the above expression, we end up with:

(1+t^2)/(2+2ti)

= (1/2)*[(1+t^2)/(1+ti)]
= (1/2)*{[(1+ti)(1-ti)]/(1+ti)}
= (1/2)*(1-ti)
= (1/2)*[1 - itan (a/2)]