Solving Complex Analysis: Find z for z^(1+4i) = i

[Tranquillity]

Hello guys!

I have to find for z in D= C \ {x in R: x<=0} with z^(1+4i) = i

a) all possible values of Log(z)
b)all possible values of z.Now my approach is:

Write z^(1+4i) = exp((1+4i) * Logz) = i = exp(i*pi/2)

which holds iff (1+4i) * Logz = i*pi/2 + i*k*2*pi where k in Z.After a lot of simplifications I find that Logz= i*pi*(1+4k) / 34 + (4*pi*(1+4k) / 34), k in Z.


So z = exp(Logz) = e^(i*pi*(1+4k)/34) * e^(4*pi*(1+4k)/34), k in ZNow to show that all these roots are in D= C \ {x in R: x<=0} I have to show that Argz is in (-pi, 0) as we have learned in lectures.

But the problem is that the term e^(i*pi*(1+4k)/34) has a term k involved and setting k=-1, 0, 1 yields three different principal arguments (-3*pi/34, pi/34 and 5*pi/34 respectively)

I think the exponential involving the imaginary unit i should not involve a k, but I did many times my calculations and cannot find actually which is the problem.Could please anyone help me?Thank you!

 

[Nehushtan]

$z^{1+4i}=i=e^{\left(\frac{\pi}2+2n\pi\right)i},\ n\in\mathbb Z$

$(1+4i)\log z=\left(\dfrac{\pi}2+2n\pi\right)i,\ n\in\mathbb Z$

$\log z=\dfrac{\left(\frac{\pi}2+2n\pi\right)i}{1+4i},\ n\in\mathbb Z$

$z=\exp\left(\dfrac{\left(\frac{\pi}2+2n\pi\right)i}{1+4i}\right),\ n\in\mathbb Z$

 

[alyafey22]

We can also use the definition of complex logarithmic function :

$$\log(z) = \ln(|z|) + i\text{arg}(z)
$$

$$\log(i) = \ln(1) + i\left(\frac{\pi}{2}+2n\pi \right)$$

$$\log(i) = i\left(\frac{\pi}{2}+2n\pi \right)$$

 

[Tranquillity]

Guys thanks for your replies but this was not what I was asking exactly!

 

[Opalg]

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After a lot of simplifications I find that $\log z= i\pi(1+4k) / 34 + 4\pi(1+4k) / 34$, $k \in Z$.
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Now to show that all these roots are in D= C \ {x in R: x<=0} I have to show that Argz is in (-pi, 0) as we have learned in lectures.

You want to find whether $z$ is on the negative real axis, or equivalently whether the imaginary part of $\log z$ is an odd multiple of $\pi.$ So you want to see whether $(1+4k)/34$ can ever be an odd integer. But in fact it can never be an integer at all, because the numerator is odd and the denominator is even. So the answer should be that $k$ can be any integer, and none of the solutions lie on the negative real axis.

 

[Tranquillity]

Basically I might have done something wrong!

I actually managed to solve the exercise by finding Logz and then had to say that my solution requires Im(Logz)=Argz to be in (-pi,pi)!

So this will restrict the values of k!

Then z=exp(...) in the cut plane Do, is valid for k in [-8,8]!

And the cut plane Do contain real numbers, so I didn't have to restrict my values to be integers,
just the Im(Logz) to be between (-pi,pi)!

Thanks again for everything :)

 

[Opalg]

I actually managed to solve the exercise by finding Logz and then had to say that my solution requires Im(Logz)=Argz to be in (-pi,pi)!

So this will restrict the values of k!

Then z=exp(...) in the cut plane Do, is valid for k in [-8,8]!

And the cut plane Do contain real numbers, so I didn't have to restrict my values to be integers,
just the Im(Logz) to be between (-pi,pi)!

I should have realized that you were using the notation Log (with a capital L) to mean the principal value of the log. That does indeed restrict the value of $k$ to lie in the range $[-8,8].$