Can someone explain to me, step by step, how to calculate all infinite values of, say,
$(1+i)^{3+4i}$?
I know how to calculate the principal value, but not how to get all infinite values...and I'm not sure how to insert the portion that gives me the other infinity values.
$\endgroup$ 25 Answers
$\begingroup$$Let (1+i)^{3+4i}=k$
Taking ln on both sides gives us $(3+4i)log_e{(1+i)}=log_ek\cdots(1)$
also $(1+i)=\sqrt{1^2+1^2}e^{\frac{i\pi}4}=\sqrt2e^{\frac{i\pi}4}\cdots(2)$
$log_e(1+i)$=$log_e$($\sqrt2e^{\frac{i\pi}4}$)
Substituting $(2)$ in $(1)$ we get
$(3+4i)(log_e\sqrt2+{\frac{i\pi}4}) = log_ek$
or $k = e^{(3+4i)(log_e\sqrt2+{\frac{i\pi}4})}$
NOTE :
GENERALISATION: To evaluate numbers of the form $(a+ib)^{c+id}$
Let $\sqrt{a^2+b^2}=r$ and argument of $a+ib$ be $\theta$
Then $(a+ib)=re^{i\theta}$ = $e^{log_e(r)+i\theta}$
Hence, $(a+ib)^{c+id}=e^{{log_e}{(r)(c+id)+i\theta}(c+id)}$
$\endgroup$ $\begingroup$When you write your complex number as an e-power, your problem boils down to taking the Log of $(1+i)$. Now that is $\ln\sqrt{2}+ \frac{i\pi}{4}$ and here it comes: + all multiples of $2i\pi$. So in your e-power you get $(3+4i) \times (\ln\sqrt{2} + \frac{i\pi}{4} + k \cdot i \cdot 2\pi)$ I would keep the answer in e-power form. You can now work it out.
$\endgroup$ $\begingroup$Let's suppose you've already defined $\log r$ for real $r > 0$, say, using Taylor series. Then given $z, \alpha \in \mathbb{C}$, you can define $$z^{\alpha} = \exp(\alpha \log z)$$ where
$$\exp(w) = \displaystyle \sum_{j=0}^{\infty} \dfrac{z^j}{j!} \qquad \text{and} \qquad \log(w) = \log |w| + i \arg(w)$$
This is not well-defined - it relies on a choice of argument, which is well-defined only up to adding multiples of $2\pi$. It's these multiples of $2\pi$ which give you new values of $z^{\alpha}$.
Explicitly, if $w$ is one value of $z^{\alpha}$, then so is $$w \cdot e^{2n \pi \alpha i}$$ for any $n \in \mathbb{Z}$.
Fun facts ensue:
- if $\alpha$ is an integer then $z^{\alpha}$ is well-defined
- if $\alpha$ is rational then $z^{\alpha}$ has finitely many values
- if $\alpha$ is pure imaginary then $z^{\alpha}$ is real (but not well-defined)
Let us find in general $w^z$ where $w$ and $z$ are complex. This expression is by definition equal to $$\exp\{z\ln w\}$$ where $\ln w$ is one of the complex logarithms of $w$. That is, it is $w'$ where $$e^{w'} = w.$$ Suppose $w = re^{i\theta}$. Then $$w' = \ln r + i\theta +2ik\pi$$ where $k$ is an arbitrary integer and the $\ln$ is the ordinary real-valued logarithm. (Since $r\ge 0$ this is well-defined everywhere except for $r=0$, in which case we are dealing with $0^z$, which really is ambiguous.)
Putting this back into the original formula we have the answer, that $$\begin{align} (re^{i\theta})^z & = \exp\{z (\ln r + i\theta + 2ik\pi)\}\tag{$\star$} \\ & =\exp\{z(\ln r + i\theta)\}\cdot \exp\{2ik\pi\cdot z\} \end{align}$$ where $k$ is an integer.
Now observe that although $(\star)$ seems to list an infinite number of solutions, they are not always distinct. For example, when $z$ is a real integer, the second factor, $\exp\{2ik\pi\cdot z\}$ part is 1 for every choice of $k$, and so can be disregarded.
To find out how many values of $(\star)$ are distinct, one needs to ask about the values of $e^{2ik\pi \cdot z}$. When $z$ has nonzero imaginary part, or is a real irrational, these are all distinct and there are an infinite family of values of $w^z$, given by different choices of $k$. But when $z$ is a real rational number with (lowest-terms) denominator $n$, there are exactly $n$ distinct values.
$\endgroup$ $\begingroup$let's to establish general formula.Suppose we have $(c+di)^{(a+bi)}$,Then we proceed as fellow
we take as $Z=(c+di)^{(a+bi)}$Taking $\log$ on both side$$ \log Z=\log(c+di)^{(a+bi)}$$$$=(a+bi)log(c+di)$$$$=(a+bi)(lnr+i\theta)$$ where$$r=\sqrt{c^{2}+d^{2}},\quad\theta=\tan^{-1}(\frac{d}{c})$$so lastly $$ Z=\exp^{(a+bi)(\ln r+i\theta)}$$
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