# Homework Help: Complex numbers

1. Jun 1, 2014

### MissP.25_5

Hello,
please I need help. I have no idea how to start this. Can someone guide me? This is not homework, I'm just studying on my own and I really don't know how to begin this.

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2. Jun 1, 2014

### benorin

probably ought to start with $\cos z = \frac{1}{2}\left(e^{iz}+e^{-iz}\right)$, substituting into the given equation we have

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

now multiply both sides by $2e^{iz}$ and collect all the terms on one side...

3. Jun 2, 2014

### MissP.25_5

Why must we multiply both sides by $2e^{iz}$ ?

4. Jun 2, 2014

### Nick O

Do it, and see what you end up with. I think you'll agree that it becomes much easier find the set of solutions for z.

5. Jun 2, 2014

Is this ok?

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6. Jun 2, 2014

### Nick O

Very close! You're just missing one part to capture the periodic nature of cosine.

7. Jun 2, 2014

### MissP.25_5

Which part? I don't understand. I thought that would be enough?

8. Jun 2, 2014

### Nick O

If cos(z) = 2, what is cos(z+2pi)?

9. Jun 2, 2014

### Nick O

I'll be honest and say that I don't know how to produce the extra term from that particular solution, but that's only because I'm missing something that someone else will probably pick up. The term is there nonetheless, which you will see (I believe) if you solve it with Euler's identity instead of u-substitution.

10. Jun 2, 2014

### CAF123

From $e^{iz} = 2\pm \sqrt{3}\Rightarrow iz = \ln (2 \pm \sqrt{3})$, you can expand the RHS, taking into consideration the fact the complex logarithm is multiple valued.

11. Jun 2, 2014

### MissP.25_5

So how should I solve this question? I guess my working is alright up to this part, right?

12. Jun 2, 2014

### lurflurf

in fact by convention the principle value is
$$\mathrm{Cos}^{-1}(2)=\imath\, \mathrm{Cosh}^{-1}(2)=\imath\, \mathrm{Ln}(2+\sqrt{3})$$
to find all solutions recall that
$$\cos(-x)=\cos(x) \\ \text{and for any integer k}\\ \cos(x +2k\, \pi)=\cos(x)$$

13. Jun 2, 2014

### CAF123

It is, the method is fine. Given $z \in \mathbb{C},$ what is $\log_{e} z$?

14. Jun 2, 2014

### MissP.25_5

I don't get that part. How do you calculate that?

15. Jun 2, 2014

### MissP.25_5

$\log_{e} z$ = loge(x+iy)

Is that correct?

16. Jun 2, 2014

### CAF123

It is not what I meant for you to write down - have you studied the complex logarithm?

17. Jun 2, 2014

### lurflurf

Same way you did.

$$\mathrm{Cos}^{-1}(2)=\imath\, \mathrm{Cosh}^{-1}(2)=\imath\, \mathrm{Ln}(2+\sqrt{3})\sim 1.316957896924816708625046347307968444026981971\imath$$
is the principal value.
That is the agreed reference value we give when we give only one value.
To find all values you do not need to deal with complex logs (though you could).
Just think back to real trigonometry.
We found one value call it P
then -P was another
then we could add 2n ∏

Once we find one solution, we then find them all
what are all solutions to
$$\cos(x)=2$$
is done the same way as
what are all solutions to
$$\cos(x)=0.2$$

18. Jun 2, 2014

### MissP.25_5

Yes, I have. But I don't know how to use that here.

19. Jun 2, 2014

### MissP.25_5

Ok, so what do I do next? Am I supposed to expand that log part?

20. Jun 2, 2014

### lurflurf

how would you do the one with 0.2?
It is the same.

If t is any solutions and k is an integer

$$2k\, \pi\pm t$$
is all the solutions because
$$\cos(2k\, \pi\pm t)=\cos(t)$$
by properties of cos(x)

21. Jun 2, 2014

### CAF123

Ok, so you know that $\log_e z = \text{Log}_e |z| + i\text{Arg}\,z + 2\pi k, \,\,k \in \mathbb{Z}$, where $\log$ is the complex log and $\text{Log}$ is the real log. Use this to obtain the full solution set of the equation $iz = \ln(2\pm \sqrt{3})$.

Alternatively, as lurflurf said, make it clear that you solved for the principal value and, using the properties of the cosine, tag on a $2\pi k.$

22. Jun 2, 2014

### MissP.25_5

Are you saying that I have to something like this?

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23. Jun 2, 2014

### MissP.25_5

But how do I find Arg(Z) here? What is the value of r?

24. Jun 2, 2014

### MissP.25_5

I'm not sure how to plug in the values. But this is my attempt.

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25. Jun 2, 2014

### lurflurf

You are over thinking
if we know the solution
$$t=\imath\, \mathrm{Ln}(2+\sqrt{3})$$
to find all solutions we consider the additive inverse and shifts by 2 k pi

$$t=2k\, \pi+\imath\, \mathrm{Ln}(2\pm \sqrt{3})=2k\, \pi\pm\imath\, \mathrm{Ln}(2+ \sqrt{3})$$