I'm trying to find how to calculate by hand a real number raised to an imaginary number, for example $3^i$, but I only find $i^i$. How could I solve on paper a real number raised to an imaginary? Is there any technique?
Asked
Active
Viewed 62 times
0
-
2$3^{i} = e^{i \ln(3)} = \cos(\ln(3)) + i \sin(\ln(3))$ – Robert Lee Sep 24 '22 at 01:39
2 Answers
1
In general $$z^w=e^{\log z^w}=e^{w\log z}$$.
Of course you need a branch of log.
So $$r^{iy}=e^{\log r^{ iy}}=e^{i y\log r}=\cos(y\ln r)+i\sin(y\ln r)$$ for $r\gt0$.
Here I used Euler's formula and properties of $e$ and $\log$.

calc ll
- 8,427
1
The standard technique for raising one number to another is extended to complex numbers. This standard technique, for example, to calculate $2^2$ consists of $$ 2=e^{\log2} $$ so $$ 2^2=(e^{\log2})^2=e^{2\log2}. $$ So using your example of $3$ and $i$, $$ 3^i=e^{i\log3}=\cos(\log3)+i\sin(\log3)\approx0.45+0.89i. $$ The case of $i^i$ $$ i^i=(e^{i\pi/2})^i=e^{-\pi/2}\approx0.21. $$

Suzu Hirose
- 11,660