Complex numbers equation

1. Sep 8, 2016

Mrencko

1. The problem statement, all variables and given/known data ask to find all the values in z to the equation to be true

2. Relevant equations

3. The attempt at a solution

this is my atemp of solution i dont know what else do, because the problem ask for z values
well i must add that i am thinking about x=0 and y= pi/2 will solve the equation but i dont know if i am right

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2. Sep 8, 2016

Dr Transport

What values of $x$ and $y$ could you come up with? Think about it systematically, $x$ first, do you have an exponential on the right hand side?

3. Sep 8, 2016

Mrencko

nope i dont have and exp in the right hand side

4. Sep 8, 2016

Staff: Mentor

What can you say about $x$ and $y$ if you know that $e^{4x} \cdot (\cos (4y)+ i \cdot \sin (4y)) = i$?

5. Sep 8, 2016

Mrencko

i can say, i need the exp to become 1, and i need the cos4y to become o also i need the sin(4y) become 1 to the equation be true, so i am thinking in x=0 and y=pi/8

6. Sep 8, 2016

Dr Transport

Sounds right......just think methodically about the problem, the solution will drop on out for you.

7. Sep 8, 2016

Mrencko

tell me please why i am wrong, or exactly what do you mean about methodically

8. Sep 8, 2016

Staff: Mentor

Is this the only solution?

9. Sep 8, 2016

Mrencko

well that is a good question, i dont know, is there a method to find out if this is the only solution?, i am taking this aproach by the eye method
z=0+pi/8

10. Sep 8, 2016

Dr Transport

No, $y = \pi / 8 \pm 2\pi$ are all the solutions.... Graph it out and you'll see the other solutions

11. Sep 8, 2016

Mrencko

i guess the other solution will be -15pi/8?

12. Sep 8, 2016

Mrencko

well the graph says that e^4z=i and z=(1/8)i(4piN+pi) nEz but i dont know what that means or how to manueally get there

13. Sep 8, 2016

Staff: Mentor

Formally you have $e^{x} \cdot (\cos (4y)+ i \cdot \sin (4y)) = |i| \cdot (a+ib)$. (The coefficient $4$ at the $x$ isn't really needed.)
Therefore you have the equations $1= |i| =e^x$ and $0 = a = \cos (4y)$ and $1 = b = \sin (4y)$ by simply comparing the three terms "length", "real part" and "imaginary part" which you found "by eye". This gave you $x=0$ and $y=\frac{1}{8} \pi$.
But $\cos$ and $\sin$ are repeating their values.

14. Sep 8, 2016

Mrencko

do you mean the natural cicles of sin and cos? i dont get it very clearly

15. Sep 8, 2016

Staff: Mentor

Yes exactly. Every full circle ($2 \pi$) you'll get the same values $\sin (4y) = 1$ and $\cos (4y) = 0$ again. So all $\frac{1}{8} \pi + 2k\pi \; (k\in \mathbb{Z})$ are solutions, which means that every full circle (backwards or forwards) you return to the point $i = (0,i)$ again.

This is important! Because it shows you, that the complex exponential function (and the logarithms, too) behave differently compared with the real exponential function. They are not unique anymore! Solutions without repetitions are called principal.

16. Sep 8, 2016

Mrencko

oh thanks i am a beginner in the complex variable and have many doubs about how its behave, many thanks, so this must have infinite ciclical solutions?

17. Sep 8, 2016

SammyS

Staff Emeritus
Actually you need $\ 4y = \pi / 2 \,,\$ but then you can add (integer) multiples of 2π to this.

So you then have: $\ 4y = \frac \pi 2 +2\pi k \,.\$ Now divide by 4

18. Sep 8, 2016

Staff: Mentor

Oops.

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