1. ## Complex Number

In order to start complex differentiation we need to understand some properties of complex numbers. However I am stuck here and would appreciate some help if possible. thanks very much

If $0 < \theta < \frac{\pi}{4}$, find the modulus and principal value of the argument of $e^{i\theta} + e^{5i \theta}$.

$|e^{i\theta} + e^{5i \theta}|^2 = (\cos \theta + \cos 5 \theta)^2 + (\sin \theta + \sin 5 \theta)^2 = 2 + 2 \cos 4 \theta$ so the modulus is $\sqrt{ 2 + 2\cos 4 \theta}$ but I don't really see how to calculate the argument and how to involve the restrictions on theta. Thanks for any help

2. This may not be very pleasant (simplified) but it is nonetheless fact.
$\text{Arg}(e^{i\theta } + e^{5i\theta } ) = \arctan \left( {\frac{{\sin (\theta ) + \sin (5\theta )}}{{\cos (\theta ) + \cos (5\theta )}}} \right)$

3. Ahh i thought about writing that but it looked horrific ! does it matter that theta is between zero and pi/4?

4. Originally Posted by slevvio
does it matter that theta is between zero and pi/4?
We do need this to be true $\cos(\theta)+\cos(5\theta)>0$.

5. Originally Posted by slevvio
In order to start complex differentiation we need to understand some properties of complex numbers. However I am stuck here and would appreciate some help if possible. thanks very much

If $0 < \theta < \frac{\pi}{4}$, find the modulus and principal value of the argument of $e^{i\theta} + e^{5i \theta}$.

$|e^{i\theta} + e^{5i \theta}|^2 = (\cos \theta + \cos 5 \theta)^2 + (\sin \theta + \sin 5 \theta)^2 = 2 + 2 \cos 4 \theta$ so the modulus is $\sqrt{ 2 + 2\cos 4 \theta}$ but I don't really see how to calculate the argument and how to involve the restrictions on theta. Thanks for any help
A little trigonometry comes in handy here:

$1+\cos 2\theta=2\cos^2\theta\,,\,\,so\,\,\,\sqrt{2+2\cos 4\theta}=2\cos 2\theta$ ...and this looks nicer, doesn't it? (**)

As for the argument: using $\sin\alpha+\sin\beta=2\sin\frac{\alpha+\beta}{2}\c os\frac{\alpha-\beta}{2}\,,\,\,\cos\alpha+\cos\beta=2\cos\frac{\a lpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$ , we get:

$\sin\theta+\sin 5\theta=2\sin 3\theta\cos 2\theta$

$\cos\theta+\cos 5\theta=2\cos 3\theta\cos 2\theta$

So for the argument we have $\arctan\left(\frac{2\sin 3\theta\cos 2\theta}{2\cos 3\theta\cos 2\theta}\right)=\arctan\left(\tan 3\theta\tan 2\theta\right)$

Tonio

6. Originally Posted by tonio
So for the argument we have $\arctan\left(\frac{2\sin 3\theta\cos 2\theta}{2\cos 3\theta\cos 2\theta}\right)=\arctan\left(\tan 3\theta\tan 2\theta\right)$
But be careful. For a principal argument using arctangent alone we need the real part to be positive.

7. $e^{i\theta} + e^{5i\theta}$

$= e^{3i\theta}( e^{-2i\theta} + e^{2i\theta})$

$= e^{3i\theta} 2 \cos(2\theta)$

the argument is $3\theta$

the modulus is $| e^{3i\theta} 2 \cos(2\theta) | = 2 |\cos(2\theta)|$