topic Re: Problem with basic calculations in Mathcad Installation

Problem with basic calculations

gstavrakis — Mon, 16 Mar 2026 10:34:56 GMT

mathcad cannot handle correctly the swap of negative symbol in odd powers.
Specific, calculating this expression (-1/8)^(-2/3) gives an imaginary result instead of the 4 as result

Re: Problem with basic calculations

terryhendicott — Mon, 16 Mar 2026 11:26:53 GMT

Prime gets this:

Maple Flow gets this:

Matlab gets this:

Julia:

Cheers

Terry

Re: Problem with basic calculations

terryhendicott — Mon, 16 Mar 2026 12:14:15 GMT

Octave:

Wolfram:

Python:

Re: Problem with basic calculations

gstavrakis — Mon, 16 Mar 2026 15:29:14 GMT

The original solution as presented in the book MODERN ENGINEERING MATHEMATICS
Glyn James, Phil **bleep**, Sixth Edition

page 7

is as shown in screenshot. Its just a simple and direct solution on this.

Re: Problem with basic calculations

terryhendicott — Mon, 16 Mar 2026 18:26:36 GMT

Hi,

Both answers are correct:

Re: Problem with basic calculations

terryhendicott — Tue, 17 Mar 2026 07:22:20 GMT

Thanks for spotting this, remember 7 mathematical programs get the complex result.

You can with some manipulation of the problem get 4.

The first step in the solution from the book gets 4.

It seems cube root gets 4 raising to the 1/3 power gets the complex result.

Very interesting learning thank you for spotting this.

Cheers

Terry

Re: Problem with basic calculations

gstavrakis — Tue, 17 Mar 2026 07:27:37 GMT

My original question is. I see two answers. One with complex number and one with simple number.

Doing simplifications i get the answer (4). A positive number.

The complex number is negative.

So, as it is just a presentation of a number, to my understanding, I will get a single answer. Otherwise, I will have to accept that every number has also a complex expression representing it. So, 4 = -2 -3.464i. I dont know much of mathematic theory if it is stands as logical, so please forgive me if I think it as not a logic answer.

If it is really accepted that each non imaginary number can have an expression as an imaginary number, then it is correct the answer to have two results. If it is not, then there is something that has to be corrected.
Mathcad gives as the answer the complex number only, when I write this number as given, as shown in first @terryhendicott 's screenshot.

Re: Problem with basic calculations

gstavrakis — Tue, 17 Mar 2026 07:31:23 GMT

Exactly that I think it is the problem with the mathematic softare. They cannot handle the swap of negative numbers when risen to 1/3 power, because those are not complex but simple numbers. (-4) ^ 1/3 is not a complex number as the negative sign swaps. They handle it correctly when i give it directly as 1/3 power but when i write this 1/3 power in a more complex way, they assume we have a complex number always as result

Re: Problem with basic calculations

ppal — Tue, 17 Mar 2026 20:45:45 GMT

Not Mathcad But Latex code:

\documentclass{article}
\usepackage{amsmath, amssymb}

\begin{document}

\section*{Rigorous Evaluation of $\left(-\frac{1}{8}\right)^{-2/3}$}

The expression involves a negative base raised to a rational power, which is multi-valued in the complex domain $\mathbb{C}$. We define the operation using the complex exponential and logarithm:
\[ f(z) = z^a = e^{a \log(z)} \]

\subsection*{1. Simplification of the Base}
First, we apply the negative exponent to find the reciprocal:
\[ \left( -\frac{1}{8} \right)^{-2/3} = \left[ \left( -\frac{1}{8} \right)^{-1} \right]^{2/3} = (-8)^{2/3} \]

\subsection*{2. Polar Representation of the Base}
The complex number $z = -8$ lies on the negative real axis. Its magnitude $r$ and argument $\theta$ are:
\[ |z| = 8, \quad \arg(z) = \pi + 2k\pi \quad \text{for } k \in \mathbb{Z} \]
Thus, in exponential form:
\[ -8 = 8e^{i(\pi + 2k\pi)} \]

\subsection*{3. Application of the Rational Exponent}
Substituting this into the power function:
\[ (-8)^{2/3} = \left( 8e^{i(\pi + 2k\pi)} \right)^{2/3} = 8^{2/3} \cdot e^{i\frac{2}{3}(\pi + 2k\pi)} \]
Since $8^{2/3} = (\sqrt[3]{8})^2 = 4$, the general solution is:
\[ w_k = 4e^{i\left( \frac{2\pi}{3} + \frac{4k\pi}{3} \right)} \]

\subsection*{4. The Three Distinct Roots}
For $n=3$ (the denominator of the exponent), there are exactly three unique values for $k = 0, 1, 2$:

\begin{itemize}
\item \textbf{Case $k=0$ (Principal Value):}
\[ w_0 = 4e^{i\frac{2\pi}{3}} = 4 \left( \cos \frac{2\pi}{3} + i \sin \frac{2\pi}{3} \right) = 4 \left( -\frac{1}{2} + i\frac{\sqrt{3}}{2} \right) = -2 + 2i\sqrt{3} \]

\item \textbf{Case $k=1$ (Real Root):}
\[ w_1 = 4e^{i\frac{6\pi}{3}} = 4e^{i2\pi} = 4(\cos 2\pi + i \sin 2\pi) = 4(1 + 0i) = 4 \]

\item \textbf{Case $k=2$ (Conjugate Root):}
\[ w_2 = 4e^{i\frac{10\pi}{3}} = 4 \left( \cos \frac{4\pi}{3} + i \sin \frac{4\pi}{3} \right) = 4 \left( -\frac{1}{2} - i\frac{\sqrt{3}}{2} \right) = -2 - 2i\sqrt{3} \]
\end{itemize}

\subsection*{Conclusion}
The "two answers" usually cited are the **Real Root ($4$)** used in elementary algebra and the **Principal Value ($-2 + 2i\sqrt{3}$)** used by computational engines like WolframAlpha.

\end{document}