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Lets say you have an equation with the following form.
$$(a+ib)^{c+id}$$

How do you go about doing it? I am very familiar with De Moivre's Theorem where d is zero and have that all set. I am unfamiliar with d being real not equal to zero.

Any additional information would be very helpful.

Will
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  • You say you have an equation, but what you presented is not an equation (it does not have an equals sign). – JRN Dec 19 '16 at 01:59

3 Answers3

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When $\exists\space\text{z}_1\space\wedge\space\exists\space\text{z}_2\in\mathbb{C}$:

$$\text{z}_1^{\text{z}_2}=\left(\left|\text{z}_1\right|e^{\left(\arg\left(\text{z}_1\right)+2\pi\text{k}_1\right)i}\right)^{\text{z}_2}=\left|\text{z}_1\right|^{\text{z}_2}e^{\text{z}_2\left(\arg\left(\text{z}_1\right)+2\pi\text{k}_1\right)i}$$

Where $\left|\text{z}_1\right|=\sqrt{\Re^2\left[\text{z}_1\right]+\Im^2\left[\text{z}_1\right]}$,$\arg\left(\text{z}_1\right)$ is the complex argument of $\text{z}_1$ and $\exists\space\text{k}_1\in\mathbb{Z}$

So:

  • $$\left|\text{z}_1\right|^{\text{z}_2}=\exp\left(\ln\left(\left|\text{z}_1\right|^{\text{z}_2}\right)\right)$$
  • $$e^{\text{z}_2\left(\arg\left(\text{z}_1\right)+2\pi\text{k}_1\right)i}=e^{\left(\Re\left[\text{z}_2\right]+\Im\left[\text{z}_2\right]i\right)\left(\arg\left(\text{z}_1\right)+2\pi\text{k}_1\right)i}=e^{\left(\Re\left[\text{z}_2\right]i-\Im\left[\text{z}_2\right]\right)\left(\arg\left(\text{z}_1\right)+2\pi\text{k}_1\right)}$$
Jan Eerland
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If $a, ,b, c, d\in\Re$ We first compute $$(a+bi)=r\times e^{i\theta}, r\neq 0$$

$\begin{eqnarray*}(a+bi)^{c+di}&=& (r\times e^{i\theta})^{c+di}\\ &=& (r^c\times e^{ci\theta})\times r^{di}\times e^{i\theta \times di}\\ &=&(r^c e^{-d\theta})\times e^{i(c\theta + d\times \ln{r})} \end{eqnarray*}\\$

Where $(r^c e^{-d\theta})$ is your radios and $e^{i(c\theta + d\times \ln{r})}$ is your unit complex number.

Ching-Ting Wu
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Let $z = a + b \mathrm{i}$. Then $z = |z| \mathrm{e}^{\mathrm{i} \arg z}$ and $$ z^{c+d \mathrm{i}} = |z|^{c+d \mathrm{i}} \mathrm{e}^{(c+d \mathrm{i})\mathrm{i} \arg z} \text{.}$$

You say you can finish from here.

Eric Towers
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  • Omitting to mention that $z^w$, when $z$ and $w$ are complex numbers, can only be multivalued is to miss the elephant in the room here. – Did Dec 10 '16 at 07:35
  • @Did: There is a difference between $\arg$ and $\mathrm{Arg}$. – Eric Towers Dec 13 '16 at 00:09
  • If a proper understanding of your answer relies on this subtlety and that you do not even mention it, then the phrasing of your answer very much sounds like intentional deception. – Did Dec 13 '16 at 09:31
  • @Did : When did nonuniqueness of argument become subtle. OP claims to know De Moivre's Theorem, $(\cos(\theta) + \mathrm{i} \sin(\theta))^n = \dots$ ($n \in \Bbb{Z}$), which is explicitly periodic in $\theta$. Treating $\arg z$ as an equivalence class mod $2\pi$ is the non-deceptive choice. – Eric Towers Dec 13 '16 at 16:00
  • Suuure... so I suppose you will explain that actually $z^w$ is a collection of complex numbers, not a complex number? Which is the reason why you do not even mention the fact and why you take case to denote it as if it was in fact a complex number? Makes sense... More seriously, when you will have decided to stop the damage control operation and to start actually addressing the comments made on the mathematics of your answer, just ring me. Otherwise... – Did Dec 13 '16 at 18:07
  • No need. Evaluation always projects away whatever non-distinction you are trying to indicate. – Eric Towers Dec 14 '16 at 17:58
  • When evaluating $z^w$ with $w$ complex, like here? Far from it. (So, I was not sure at the beginning because of its ambiguous formulation, but now it is clear, this answer is quite wrong.) – Did Dec 14 '16 at 20:38
  • @Did : I notice that you have sniped without substance at more than one answer here but cannot be bothered to submit an answer you would find to be correct nor to demonstrate a fault. – Eric Towers Dec 15 '16 at 20:18
  • Irrelevant. In actual life, I pointed specifically at the problem with your answer in my very first comment, problem which is well known in complex analysis,and since then we are witnessing your excuses to avoid addressing the correctness of your post (which should be your main concern, but alas...). Do you still think that "Evaluation always projects away (the) distinction (I am) indicat(ing)", whatever this imprecise sentence may mean? To get minimally educated on the subject, you might start by scanning all the questions on the site showing your assertion is absurd. – Did Dec 15 '16 at 21:25
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    @Did : You snipe at Ching-Ting Wu's answer for missing multivaluedness. You take multiple tries to snipe at my answer for having multivaluedness. Since your arguments are manifestly incoherent, you should not be surprised that no one is paying any attention at all. – Eric Towers Dec 15 '16 at 21:48
  • "your arguments are manifestly incoherent" Yes? Care to provide a mathematical justification for this (startling) assertion? (Note: 1. You still focus on irrelevant facts that interest you only and show you are simply not interested in mathematical correctness. 2. Using "snipe" merely signals that you do not understand the goal of the site.) – Did Dec 15 '16 at 21:52
  • Do you still think that "Evaluation always projects away (the) distinction (I am) indicat(ing)"? To make it simple, do you still think that $e^{(c+id)i\theta}$ is the same for every $\theta$ such that $e^{i\theta}$ is a given complex number? – Did Dec 15 '16 at 21:53