Use Euler’s formula to write an exponential function in the form $a + ib$
By Gabriel Cooper •
Euler's formula is $e^{it} = \cos(t) + i \sin(t)$.
$$2^{1-i} = 2\left(2^{-i}\right)=2\cos(-1) + 2\ i\sin(-1)$$ $$\pi^{-1+2i} = \pi^{-1}\left(\pi^{2i}\right)=\pi^{-1}\left(\cos 2+i\sin 2\right)=\frac{\cos 2}{\pi}+i\ \frac \sin 2 \pi$$
Is the above the correct way to write $2^{1-i}$ and $\pi^{-1+2i}$ in the form $a + ib$? I wasn't sure if I could change Euler's formula to $n^{it} = \cos(t) + i \sin(t)$, where $n$ is a constant.
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$\begingroup$No. It is not true that $n^{it} = \cos(t) + i \sin(t)$. Notice that $$ 1^{it} = 1 \neq \cos(t) + i \sin(t), $$ for example consider this at $t=0$.
As a hint for figuring this out, notice that $$ x^{a+ib} = e^{\ln(x^{a+ib})} $$ then recall your rules for logarithms to get this to the form $e^{(a+ib)\ln(x)}$.
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