Find the derivative of $z^{\sin z}$ at $z = i$

Question

So I have encountered a question in which we have to evaluate the derivative of $$z^{\sin z}$$ at $z = i$ using the principal branch of the complex power that is using the principal branch of log (Log).
I have tried expanding the complex power using the formula $$z^{\sin z}=e^{(\sin z)\operatorname{Log}(z)}$$ To obtain $u(x,y)$ and $v(x,y)$ but this yields a lengthy-expression which isn't very easy to differentiate. Is there a way to differentiate this easily or do we have to differentiate the expanded complex power and then substitute z = i?

Answer 1

$$(z^{sinz})' = (e^{sinz \cdot lnz})' = (cosz \cdot lnz + sinz \cdot 1/z)\cdot e^{sinz \cdot ln z} = (cosz \cdot lnz + sinz \cdot 1/z) z^{sinz}$$

Since $ \sin(ix) = i \sinh(x) $ and $ \cos(ix) = cosh(x) $:

$$(z^{sinz})'|_{z=1} = (\cosh(1) \cdot (0+ i (\pi / 2+2k\pi)) + i \sinh(1) \cdot 1/i) \cdot i^{i \sinh(1) }$$

Answer 2

It helps to use logarithms and implicit differentiation.

$$u := z^{\sin z}$$ $$\log u = \sin z \log z$$ $$\frac{d}{dz} \log u = \frac{d}{dz}(\sin z \log z)$$ $$\frac{1}{u} \frac{du}{dz} = (\log z)(\frac{d}{dz}\sin z) + (\sin z)(\frac{d}{dz} \log z)$$ $$\frac{1}{z^{\sin z}} \frac{du}{dz} = (\log z)(\cos z) + (\sin z)(\frac{1}{z})$$ $$\frac{du}{dz} = \frac{z^{\sin z}(z(\log z)(\cos z) + \sin z)}{z}$$

Now, what do you get when you plug in $z = i$? The identity $e^{ix} = \cos{x} + i\sin{x}$ may be useful.

Answer 3

You can show with logarithmic differentiation that $$f(z)=\frac{\mathrm d}{\mathrm dz}(z^{\sin z})=z^{-1+\sin z}(\sin z+z\log z\cos z)$$

So $$f(\mathrm i)=\mathrm i^{-1+\sin \mathrm i}(\sin \mathrm i+\mathrm i\log \mathrm i\cos \mathrm i)$$

We can compute this. See, $$\sin(z)=\frac{\mathrm e^{\mathrm iz}-\mathrm e^{-\mathrm iz}}{2\mathrm i} \\ \sin(\mathrm i)=\frac{1}{2\mathrm i}(\mathrm e^{-1}-\mathrm e)$$ Similarly $\cos(\mathrm i)=\frac{1}{2}(\mathrm e^{-1}+\mathrm e)$.

Recall also that $$\log z=\log|z|+\mathrm i\operatorname{arg}z$$ So $$\log \mathrm i=\log|\mathrm i|+\mathrm i\operatorname{arg}\mathrm i=\frac{\pi}{2}\mathrm i$$ Finally, $$\mathrm i^s=\exp(s\log\mathrm i)=\exp\left(s~\frac{\pi}{2}\mathrm i\right)$$ So $$\mathrm i^{-1+\sin \mathrm i}=\exp\left((-1+\sin \mathrm i)~\frac{\pi}{2}\mathrm i\right) \\ =\exp\left(\left(-1+\frac{1}{2\mathrm i}(\mathrm e^{-1}-\mathrm e)\right)~\frac{\pi}{2}\mathrm i\right) \\ =\exp\left(-\frac{\pi}{2}\mathrm i+\frac{\pi}{4}(\mathrm e^{-1}-\mathrm e)\right) \\ =\mathrm e^{-\mathrm i\pi/2}\mathrm e^{\pi(\mathrm e^{-1}-\mathrm e)/4}$$

Putting it all together, $$f(\mathrm i)=\mathrm e^{-\mathrm i\pi/2}\mathrm e^{\pi(\mathrm e^{-1}-\mathrm e)/4}~\left(\frac{1}{2\mathrm i}(\mathrm e^{-1}-\mathrm e)+\mathrm i~\frac{\pi}{2}\mathrm i~\frac{1}{2}(\mathrm e^{-1}+\mathrm e)\right) $$ Simplifying and using $\mathrm e^{-\mathrm i\pi/2}=-\mathrm i$, $$f(\mathrm i)=-\mathrm i~\mathrm e^{\pi(\mathrm e^{-1}-\mathrm e)/4}\left(\frac{1}{2\mathrm i}(\mathrm e^{-1}-\mathrm e)-\frac{\pi}{4}(\mathrm e^{-1}+\mathrm e)\right) \\ =\mathrm e^{\pi(\mathrm e^{-1}-\mathrm e)/4}\left(\frac{\mathrm i\pi}{4}(\mathrm e^{-1}+\mathrm e)+\frac{1}{2}(\mathrm e-\mathrm e^{-1})\right)$$

This has an approximate numerical value of $$0.185526+0.382649~\mathrm i$$

As confirmed by Mathematica: