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[Pouyan]

Consider the principal branch of the function

f(z)= z^7/3

Find f'(-i) and write it in the form a+bi

My attemp is :

I know z^c = exp(c logz)
and the derivative of that is : (c/z) * exp(c Logz)
That is in this case (7/3)*(i) *exp((7/3)*Log-i) = f'(-i)
I know that Log(-i) = Log(1) + i(-pi/2)= -i pi/2
and exp((7/3)(-i pi/2)) = (cos(7pi/6)-i*sin(7pi/6))
and I get (7/3)*(i)(cos(7pi/6)-i*sin(7pi/6)) as final answer

But I see the answer is : -(7/6)(1+isqrt(3))!
What is wrong with my algorithm ?

 

[ehild]

Consider the principal branch of the function

f(z)= z^7/3

Find f'(-i) and write it in the form a+bi

My attemp is :

I know z^c = exp(c logz)
and the derivative of that is : (c/z) * exp(c Logz)
That is in this case (7/3)*(i) *exp((7/3)*Log-i) = f'(-i)
I know that Log(-i) = Log(1) + i(-pi/2)= -i pi/2
and exp((7/3)(-i pi/2)) = (cos(7pi/6)-i*sin(7pi/6))
and I get (7/3)*(i)(cos(7pi/6)-i*sin(7pi/6)) as final answer

But I see the answer is : -(7/6)(1+sqrt(3))!
What is wrong with my algorithm ?

Include (i) into the exponent.

 

[Pouyan]

Include (i) into the exponent.

-i is in the exponent

 

[Mark44]

Consider the principal branch of the function

f(z)= z^7/3

Find f'(-i) and write it in the form a+bi

My attemp is :

I know z^c = exp(c logz)
and the derivative of that is : (c/z) * exp(c Logz)

Is there some reason you can't use the derivative formula $\frac d {dz} z^k = k z^{k - 1}$?
Then $f'(-i) = \frac 7 3 (-i)^{4/3} = \frac 7 3 (e^{-i \pi/2})^{4/3}$
Simplify this last expression to get to the form in your textbook.

That is in this case (7/3)*(i) *exp((7/3)*Log-i) = f'(-i)
I know that Log(-i) = Log(1) + i(-pi/2)= -i pi/2
and exp((7/3)(-i pi/2)) = (cos(7pi/6)-i*sin(7pi/6))
and I get (7/3)*(i)(cos(7pi/6)-i*sin(7pi/6)) as final answer

But I see the answer is : -(7/6)(1+sqrt(3))!
What is wrong with my algorithm ?

 

[Ray Vickson]

-i is in the exponent

Easier: if $f(z) = z^n$ then $f'(z) = n z^{n-1}$. Apply that to $n = 7/3$ and $z = -i$.

Anyway,
$$\exp((7/3)(-i \pi/2)) = \cos(7\pi/6)-i \sin(7\pi/6) =- \frac{\sqrt{3}}{2} + \frac{1}{2} i $$.

 

[Mark44]

Easier: if $f(z) = z^n$ then $f'(z) = n z^{n-1}$. Apply that to $n = 7/3$ and $z = -i$.

That's what I said.

Anyway,
$$\exp((7/3)(-i \pi/2)) = \cos(7\pi/6)-i \sin(7\pi/6) =- \frac{\sqrt{3}}{2} + \frac{1}{2} i $$.

 

[ehild]

-i is in the exponent

i is also a factor:

and I get (7/3)*(i)(cos(7pi/6)-i*sin(7pi/6)) as final answer

Write the multiplicative i in exponential form and collect the exponents.

 

[Pouyan]

Thank you so much for your help. I actually mixed up this problem with the formula

Z^m/n = exp((m/n)Log|z|)exp(i(m/n)(Arg z + 2kpi))
but now I see I could solve it easier

 

[Ray Vickson]

That's what I said.

For some reason, your message did not appear on my screen until after I pressed the "enter" key. That sort of thing happens to me quite often, and I have no idea why.

In fact, I did not see your message until I had logged off and later, logged on again!