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The question asks:

Find integers $p$ and $q$ such that $(p + qj)^{5} = 4 + 4j$

The question prior to this was:

Find the fifth roots of $4 + 4j$ in the form $re^{j\theta }$, where $r > 0$ and $-\pi < \theta \leq \pi$. Illustrate these fifth roots on an Argand diagram

I am able to do that but I am unsure which root to use to answer the original question. Help would be much appreciated

Johnmgee
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  • It's definitely easiest to convert $4+4i$ (or at least I assume it must be $i$) into $re^{i\theta}$ form. Taking the fifth root of that is straightforward enough. Then you can convert back to $x+yi$ form. – Ian Coley May 04 '13 at 19:30

3 Answers3

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You say you found all five roots of the equation, presumably in exponential form. Now write them in trigonometric form and simplify. I'm guessing there will only be one case in which both the $\cos \theta$ and $\sin \theta$ will be integers.

Spine Feast
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  • Would it be more obvious given the argand diagram of the roots? I mean it is also obvious which angles give integers when multiplied by $\sqrt{2}$ from basic knowledge of the sines and cosines of the angles but I like argand diagrams :) – Johnmgee May 04 '13 at 19:34
  • Ah, you're right. Since you've already drawn it on the diagram then of course you should see straight away the required integers. And you're right, especially with complex numbers a picture is worth a thousand words (numbers) :) – Spine Feast May 04 '13 at 19:46
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$$4+4i=4\sqrt2\left(\cos\frac\pi4+i\sin\frac\pi4\right)=(\sqrt2)^4\sqrt2e^{i\frac\pi4}$$ (using Euler's Identity)

$$\implies 4+4i=(\sqrt2)^5e^{2ni\pi+\frac{i\pi}4}=(\sqrt2)^5e^{\frac{i(8n+1)\pi}4}$$ where $n$ is any integer

So, $$(4+4i)^\frac15=\sqrt2e^{\frac{(8n+1)i\pi}{20}}=\sqrt2\left(\cos \frac{(8n+1)\pi}{20}+i\sin \frac{(8n+1)\pi}{20}\right)$$ where $0\le n\le 4$

Please find the explanation here regarding the values of $n$

We need $\sqrt2$ at the denominator of $\cos,\sin$ to make $p,q$ integers

Trying with $n=0,1,2,3,4$ we find

for $n=3,\cos \frac{(8n+1)\pi}{20}+i\sin \frac{(8n+1)\pi}{20}$ becomes $\cos \frac{25\pi}{20}+i\sin \frac{25\pi}{20}=-\frac{1+i}{\sqrt2}$

$\implies$ one of the vaues of $(4+4i)^\frac15$ is $\sqrt2\left(-\frac{1+i}{\sqrt2}\right)=-(1+i)$

Community
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lab bhattacharjee
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$(\sqrt{p^2+q^2})^5=4\sqrt{2}$, so $p^2+q^2=2$, then $(p,q)=(\pm 1, \pm 1)$. It remains to check the argument $ 1\times 5, 3\times 5, -3\times 5, -1\times 5$ which one $\equiv 1\mod 8$, the answer is $(p,q)=(-1, -1)$.

Ma Ming
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