Xmas Maths 2017: The value of a $\color{red}m^{\color{green}{ince}\ \color{orange}{\pi}}$ and $\color{red}\pi^{\color{green}{ie}}$

Question

In the spirit of the festive season!

(i) Evaluate

$$\large \color{red}m^{\color{green}i\color{orange}n\color{red}c\color{green}e\;\color{red}\pi}$$

given that

$m=12\;\;\text{(month when mince pies are consumed)}\\ n=5\;\;\text{(number of points of the star on a mince pie)}\\ c=2.997\times 10^{8}s^{-1}\; \text{(rate at which mince pies are consumed!)}\\ $

(ii) It is well known that $e^{i\pi}=-1$, but what is the value of

$$\large\color{red}\pi^{\color{green}{ie}}$$

?

Merry Christmas!

Answer 1

First part: $$y=e^{ie\cdot nc\pi\cdot\ln{m}}$$ $$y=e^{i\cdot(e\cdot nc\pi\cdot\ln{m})}$$ Recognizing Euler's polar form for complex form: $$y=\cos(e\cdot nc\pi\cdot\ln{m})+i\cdot \sin(e\cdot nc\pi\cdot\ln{m})$$

Second part:

$$y=e^{ie\cdot\ln{\pi}}$$ $$y=e^{i(e\cdot\ln{\pi})}$$ Recognizing Euler's polar form for complex form: $$y=\cos{(e\ln\pi)}+i\cdot\sin{(e\ln\pi)}$$

Now you can eat this $\pi^{ie}$ ;)

Answer 2

[A Partial Answer]

Part $2$:

$$\pi^{ie} = e^{ie \log \pi} = \cos(e\log \pi) + i\sin(e\log \pi) \approx \color{red}{-1} + 0.03i$$

Answer 3

i) $$m^{ince\pi}=e^{ince\pi\ln m}=\cos(\pi nce\ln m)+i\sin(\pi nce \ln m)$$ Now simply plug the values in and evaluate :)

ii) Similarly$$\pi^{ie}=e^{ie\ln\pi}=\cos(e\ln\pi)+i\sin(e\ln\pi)=-0.99955...+0.02988...i$$ which is actually quite close to $e^{i\pi}$.

Answer 4

$$\color{blue}{\text{ii}}$$

$$\large\color{red}\pi^{\color{green}{ie}}=\left(\color{green}e^{\ln(\color{red}\pi)}\right)^{\color{green}{ie}}=\left(\color{green}e^{\color{green}{i}\color{red}{\pi}}\right)^{\frac{\color{green}{e\color{black}{\ln(}\color{red}\pi\color{black})}}{\color{red}\pi}}=\big(-1\big)^{\frac{\color{green}{e\color{black}{\ln(}\color{red}\pi\color{black})}}{\color{red}\pi}}.$$