Can I solve this complex numbers equation? Finding values for z

[Mrencko]

Homework Statement

ask to find all the values in z to the equation to be true[/B]

Homework Equations

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The Attempt at a Solution

this is my atemp of solution i don't 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 don't know if i am right

 

[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?

 

[Mrencko]

nope i don't have and exp in the right hand side

 

[fresh_42]

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

 

[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

 

[Dr Transport]

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

 

[Mrencko]

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

 

[fresh_42]

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

Is this the only solution?

 

[Mrencko]

well that is a good question, i don't know, is there a method to find out if this is the only solution?, i am taking this approach by the eye method
z=0+pi/8

 

[Dr Transport]

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

 

[Mrencko]

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

 

[Mrencko]

well the graph says that e^4z=i and z=(1/8)i(4piN+pi) nEz but i don't know what that means or how to manueally get there

 

[fresh_42]

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.

 

[Mrencko]

But coscos\cos and sinsin\sin are repeating their values.

do you mean the natural cicles of sin and cos? i don't get it very clearly

 

[fresh_42]

do you mean the natural cicles of sin and cos? i don't get it very clearly

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.

 

[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?

 

[SammyS]

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

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

 

[fresh_42]

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 and see what values of k give results for y in the interval [0, 2π) .

Oops.