Trigometric addition of complex numbers

[Mathman23]

Hello

I got some funny idears don't know if they are true, but I will share them with you.

Homework Statement

I would like to prove these formula
(1)$$sin(z_1 + z_2) = sin(z_1) \cdot cos(z_2) + sin(z_2) \cdot cos(z_1)$$

(2)$$cos(z_1 + z_2) = sin(z_1) \cdot sin(z_2) - cos(z_2) \cdot cos(z_1)$$

(3)$$cos(z)^2 + sin(z)^2 = 1$$

for any complex number $$z_1, z_2, z \in \mathbb{C}$$

The Attempt at a Solution

Proof for (1)

(step a)Let $$z_1 = x_1 + i \cdot y_1$$ and $$z_2 = x_2 + i \cdot y_2$$

(step b)According to the rules of addition of complex numbers

$$z_1 + z_2 = (x_1 + x_2) + i(y_1 + y_2)$$, where $$z_1 + z_2 \in \mathbb{C}$$

(step c)I denote $$x_1 + x_2 = x$$ and $$y_1 + y_2 = y$$

(step d)thus by replacing x and y by their respective polar coordinant representation:

$$z_1 + z_2 = (r \cdot cos(\theta)) + i \cdot ( r \cdot sin(\theta))$$, where $$z_1 + z_2 \in \mathbb{C}$$

(step e)which can be reformulated to:

$$z_1 + z_2 = r \cdot (cos(\theta)) + i \cdot ( sin(\theta))$$.

(step f)Which equals according to the euler identity:

$$z_1 + z_2 = r \cdot e^{i \theta}$$.

(step g)Then I apply sinus on both sides of the equality and get

$$\begin{array}{ccc}sin(z_1 + z_2) = cosh(sin(\theta) \cdot r) \cdot sin(cos(\theta) \cdot r) + sinh(sin(\theta) \cdot r) \cdot cos(cos(\theta) \cdot r) \cdot i \end{array}$$

From the above we see that $$x = r \cdot cos(\theta)$$ and $$y = r \cdot sin(\theta)$$

(step h)
With this in mind we rearrange the above and remembers the identity for x and y.

$$sin(z_1 + z_2) = sin(x_1 + x_2) \cdot cosh(y_1 + y_2) + cos(x_1 + x_2) \cdot sinh(y_1 + y_2) \cdot i$$

I have thus proved the sinus addition formula for complex numbers. (q.e.d)

(2) Proof

The steps are the same until (step g) where I take cosine on both sides of the equality and then obtain

$$cos(z_1 + z_2) = cos(x_1 + x_2) \cdot cosh(y_1 + y_2) - sin(x_1 + x_2) \cdot sinh(y_1 + y_2) \cdot i$$

q.e.d.

Test for (1): Let $$z_1 = 2 + 2i$$ and $$z_2 = 4 + 2i$$

Then by applying my formula on

$$z_1 + z_2 = 6 + 4i$$

$$sin(z_1 + z_2) = sin(6 + 4i) = sin(6) \cdot cosh(4) + cos(6) \cdot sinh(4) \cdot i$$

Thus my formula is true.

Test for (1): Let $$z_1 = 2 + 2i$$ and $$z_2 = 4 + 2i$$

Then by applying my formula on

$$z_1 + z_2 = 6 + 4i$$

$$cos(z_1 + z_2) = cos(6 + 4i) = cos(6) \cdot cosh(4) - sin(6) \cdot sinh(4) \cdot i$$

Thus my formula is true.


Best Regards

Fred

Homework Equations

p.s. Does my proof look okay? If yes could somebody please give a hint for (3)??

 

[Gib Z]

I haven't looked through the 2 proofs very well, but they look fine. For 3, Write sin x and cos x in exponential form from Eulers Formula. Write z = a+bi, and expand.

 

[Mathman23]

Hi Gib and Dick,

Thank You for Your answer,

Regardin (3)

Proof:

Let $$z = a +ib$$ be a complex number.

Then according to Eulers formula:

$$sin(z) = \frac{e^{iz} - e^{-iz}}{2i}$$

$$cos(z) = \frac{e^{iz} + e^{-iz}}{2}$$

Then by squaring the above and adding them together (My motovation for doing this part, is that that the point (sin(z),cos(z)) lies on the unit cicle (x^2 + y^2 = 1) ?)

$$(\frac{e^{iz} - e^{-iz}}{2i})^2 + (\frac{e^{iz} + e^{-iz}}{2})^2 =1$$

Thus the formula for $$sin(z)^2 + cos(z)^2 =1$$

where z is complex number is true. (q.e.d)

Best Regards
Fred

p.s. You you please be so kind to comment my proofs for (1) and (2) :)

p.p.s. Thank You Dick for you correction of my trig.add mistakes. It was a dumb a** mistake on my part.

 

[Dick]

Square the exponential expressions and add them. cos(z) equals negative cos(a+ib)!? Don't be silly.

 

[Mathman23]

Hi Dick,

Please look at my edited post #3.

Best Regards
Fred

 

[Dick]

Hi Dick,

Please look at my edited post #3.

Best Regards
Fred

Looks fine. Did you actually expand the squares and use the rules of exponentials to verify the sum is one?

 

[Mathman23]

Looks fine. Did you actually expand the squares and use the rules of exponentials to verify the sum is one?

No I didn't.

I will just look it up.

 

[Dick]

NO. Like this:

$$(\frac{e^{iz} + e^{-iz}}{2})^2+(\frac{e^{iz} - e^{-iz}}{2i})^2 = ?? = 1$$

Where did that crazy sum come from?

 

[Mathman23]

Dear Dick

It was a cut and past error on my part , it should have said

(sin(z))^2 + (cos(z))^2 = 1.

Best regards

Fred

p.s. You mean by expand the above I write them as the sum of their squared respective power series?

 

[Dick]

No again. (cos(z))^2=((exp(iz)+exp(-iz))/2)^2=exp(2iz)/4+1/2+exp(-2iz)/4. Do you see why? What's sin(z)^2? Add them.

 

[Mathman23]

No again. (cos(z))^2=((exp(iz)+exp(-iz))/2)^2=exp(2iz)/4+1/2+exp(-2iz)/4. Do you see why? What's sin(z)^2? Add them.

Hello Dick,

I forgot to be carefull. Sorry.

By repeating you thema I get sin(z)^2


$$sin(z)^2 = (\frac{e^{iz}-e^{-iz}}{2i})^2 = \frac{-(e^{iz})^2}{4} + \frac{1}{2} - \frac{(e^{-iz})^2}{4}$$

Therefore sin(z)^2 + cos(z)^2 = 1.

Best Regards
Fred

 

[Dick]

And adding them gives you?

 

[Mathman23]

And adding them gives you?

I changed my post #11 above.

How does the proof of (3) look now? Am I steppin in the right direction?

Best Regards.

/Fred

 

[Gib Z]

Looks like the proof is complete. Well done, I guess there was no need to let z=a+bi after all.

 

[Mathman23]

Hi Just a clean write of part 3,

Proof (Part 3):

Let $$z$$ be a complex number.

Then according to Eulers formula:

$$sin(z) = \frac{e^{iz} - e^{-iz}}{2i}$$

$$cos(z) = \frac{e^{iz} + e^{-iz}}{2}$$


Its know that

$$(sin(z))^2 = (\frac{e^{iz} - e^{-iz}}{2i})^2 = -\frac{(e^{iz})^2}{4} + \frac{1}{2} - \frac{(e^{-iz})^2}{4}$$

$$(cos(z))^2 = (\frac{e^{iz} + e^{-iz}}{2})^2 = \frac{(e^{iz})^2}{4} + \frac{1}{2} + \frac{(e^{-iz})^2}{4}$$

By addin them together I get

$$(sin(z))^2 + (cos(z))^2 = (-\frac{(e^{iz})^2}{4} + \frac{1}{2} - \frac{(e^{-iz})^2}{4}) + (\frac{(e^{iz})^2}{4} + \frac{1}{2} + \frac{(e^{-iz})^2}{4}) = 1$$

Thus the formula for $$sin(z)^2 + cos(z)^2 = 1$$

where z is complex number is true. (q.e.d)

Best Regards
Fred

p.s. Could somebody please be so kind and look through this part 3, and see if I have overlooked anything?

 

[Dick]

Looks good to me.

 

[Mathman23]

Looks good to me.

Regarding (1) and (2) are they okay too?

Any need to try to factor my formula result so they look like the original formulas

sin(z1 + z2) = sin(z1) * cos(z2) + sin(z2) * cos(z1) ?

or is my way of formulating it consistent??

Best Regards
Fred

 

[Dick]

(1) and (2) don't look so great. Just do them the same way you did (3). Replace all of the cos's and sin's by exponentials on both sides of the equations. Then multiply everything out and show both sides are equal. BTW you have the two terms on the RHS reversed in your statement of (2).

 

[Mathman23]

Okay thanks,

Thought I was too lucky about 1 and 2.

by using the euler formula for sin(z1+z2)

I get following:

$$\begin{array}{cccc}sin(z_1+z_2) &=& \frac{e^{i \cdot z_1} + e^{- i \cdot z_1}}{2} \cdot \frac{e^{i \cdot z_2} - e^{- i \cdot z_2}}{2i} + \frac{e^{i \cdot z_2} + e^{- i \cdot z_2}}{2} \cdot \frac{e^{i \cdot z_1} - e^{- i \cdot z_1}}{2i} \\ &=& sin(z_1) \cdot cos(z_2) + sin(z_2) \cdot cos(z_1) \end{array}$$


By using euler on cos(z_1 + z_2)

$$\begin{array}{cccc}cos(z_1+z_2) &=& \frac{e^{i \cdot z_2} - e^{- i \cdot z_2}}{2i} \cdot \frac{e^{i \cdot z_1} - e^{- i \cdot z_1}}{2i} + \frac{e^{i \cdot z_2} + e^{- i \cdot z_2}}{2} \cdot \frac{e^{i \cdot z_1} + e^{- i \cdot z_1}}{2} \\ &=& cos(z_1) \cdot cos(z_2) - sin(z_1) \cdot sin(z_2) \end{array}$$


Is that what you mean?

Best regards

Fred

 

[Gib Z]

Yes that Looks excellent. Good work.

 

[Mathman23]

Hi again,

I have been lookin through (2) again, and I cannot get it to work using euler.

I have found in my textbook

that exp(z1 + z2) = exp(z1) * exp(z2)

Therefore

let $$f(z) = sin(z) \cdot cos(z-c)$$ where z and c belongs to $$\mathbb{C}$$

Then

$$f'(z) = cos(z) \cdot cos(z-c) - sin(z) \cdot sin(z-c)=0$$

The only way fo this to work is if f'(z) = f'(0)

Thusly friends...

$$cos(c) = cos(z) \cdot cos(z-c) - sin(z) \cdot sin(z-c)$$

and if $$z = z_1$$ and $$c = z_1 + z_2$$

$$cos(z_1 + z_2) = cos(z_1) \cdot cos(z_2) - sin(z_1) \cdot sin(z_2)$$

I my argument waterproof here? If yes am I missing something ?

/Fred

 

[Dick]

I thought what you were trying to do was use the exponential definition of sin and cos to prove trig identities. If so then just multiply out the first exponential expression in post #19 and show that it's equal to (exp(i(z1+z2))-exp(i(z1+z2)))/2i, which is sin(z1+z2).

 

[Mathman23]

My post above regardin (2), was just an experiment, but I am unable to argue for

f'(z) = f'(0) :(


But anyway using post 19 again.

how am suppose to get the minus between

cos(z_1) * cos(z_2) - sin(z_1) * sin(z_2) = cos(z1 + z_2)

Because I multiply by Eulers formula. I get

cos(z_1) * cos(z_2) + sin(z_1) * sin(z_2) = cos(z1 + z_2) ??

/Fred

 

[Dick]

My post above regardin (2), was just an experiment, but I am unable to argue for

f'(z) = f'(0) :(


But anyway using post 19 again.

how am suppose to get the minus between

cos(z_1) * cos(z_2) - sin(z_1) * sin(z_2) = cos(z1 + z_2)

Because I multiply by Eulers formula. I get

cos(z_1) * cos(z_2) + sin(z_1) * sin(z_2) = cos(z1 + z_2) ??

/Fred

No you don't. You get a minus sign. Did you forget i*i=(-1)? If you do this carefully, I really promise you it will work.

 

[Mathman23]

thanks man it did :)

Best regards

Fred