Converting Complex Numbers to Trigonometric Identities

[Joza]

I was never good at trigonometric identities.


Let z= cos x + i sin x

Express 2/(1 + z) in the form 1 - i tan kx



I need help. A pointer to where to start would be great.

 

[tim_lou]

$$z=e^{ix}$$

and
$$\tan(kx)=\frac{1}{2i}\cdot\frac{e^{ikx}-e^{-ikx}}{e^{ikx}+e^{ikx}}$$

 

[Joza]

I'm sorry, but I have never seen that identity before. It's not in my math tables?

 

[HallsofIvy]

Do you know that $tan z= \frac{sin z}{cos z}$? Replace sin z with $(e^{ix}-e^{-ix})/2i$ and cos z with $(e^{ix}+ e^{-ix})/2$

I would have been inclined to do this problem more directly. Write 2/(1+ z) as
$$\frac{2}{1+ cos x+ i sin x}= \frac{2}{(1+cos x)+ i sin x}$$
and multiply both numerator and denominator by the "complex conjugate", (1+ cos x)- i sin x.

 

[Gib Z]

I'm sue Joza hasnt seen polar form yet...HallsofIvys method is the best way for you.

 

[Joza]

No I understand polar form. And yes I understand that tan is sin/cos. I haven't done 3rd level mathematics yet though.

 

[Gib Z]

O well that's good then, Halls's method is still the way to go though.

 

[uart]

O well that's good then, Halls's method is still the way to go though.

That seemed the most obvious way to proceed to me as well. I was going to post exactly that but Tim beat me to it, then I decided I liked his way better.

Here's the reason. If you follow through with the "conjugate" method and simplify as much as possible you end up with,

$$\frac{2}{1+z} = 1 - i \, \frac{\sin(x)}{1+\cos(x)}$$.

While this is tantalizingly close to the form that OP requires it's still not quite there. To finish off you need to be able the see that you can use the following two trig identities (on the numerator and denominator respectively),

$$\sin(x) = 2 \sin(x/2) \cos(x/2)$$

and

$$1 + \cos(x) = 2 \cos^2(x/2)$$.

This last step wasn't too obvious to me, actually I didn't even see it at first. It was only after working through Tim’s method that I realized that "k" had to be equal to 1/2 and then the half angle substitutions idea came to me.

 

[tim_lou]

actually... just look at
$$1-\frac{2}{z+1}=\frac{z-1}{z+1}$$ alone with the identity will give you everything.