Re: HSC 2016 4U Marathon
![](https://latex.codecogs.com/png.latex?\bg_white $Note however that if $z = r\, \mathrm{cis} (\theta)$ is a general non-zero complex number and $\alpha = \frac{1}{n}$ for some positive integer $n$ (where $r,\theta \in \mathbb{R}$), then $z^{\alpha}$ will be \textsl{multivalued}. This means that we need to assign a principal value to the symbol $z^\alpha$, like how for positive real numbers, we have assigned a principal value for $x^{\frac{1}{2}}$ (namely we take that symbol to mean the \textsl{positive} square root of $x$). In general, it is not true that $\mathcal {PV}\left( z^\alpha \right)$ (where $\mathcal{PV}$ stands for principal value) will be the same as $r^\alpha \mathrm{cis} \left (\alpha \theta\right)$. However, it is true that $r^\alpha \mathrm{cis} \left (\alpha \theta\right)$ will be one of the values of the multivalued $z^\alpha$ (in fact this also is true for general non-real complex exponents $\alpha$, but non-real exponents are definitely not in the syllabus).$)
![](https://latex.codecogs.com/png.latex?\bg_white $So in general, if a complex number is raised to a non-integer power, the result is multivalued, but using De Moivre's Theorem always gets you one of the values.$)
That's just the case when n=1/2 (NOTE: N ACTUALLY DOES NOT HAVE TO BE AN INTEGER. It's only called De Moivre's theorem when we deal with positive integers that's all. the statement is actually true for all real n)
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