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#### Tranquillity

##### Member

- Feb 22, 2012

- 36

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 Z

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.

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!