If $z^4 + \dfrac {1}{z^4}=47$ then find the value of $z^3+\dfrac {1}{z^3}$
My Attempt: $$z^4 + \dfrac {1}{z^4}=47$$ $$(z^2+\dfrac {1}{z^2})^2 - 2=47$$ $$(z^2 + \dfrac {1}{z^2})^2=49$$ $$z^2 + \dfrac {1}{z^2}=7$$
How do I proceed further??
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Z is complex number? – Khosrotash Jun 25 '17 at 13:33
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@Khosrotash It's algebra-precalculus. – Michael Rozenberg Jun 25 '17 at 13:34
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@Khosrotash, no.. that's not – pi-π Jun 25 '17 at 13:38
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Look at for example this, this or this or some of the other earlier variants. – Jyrki Lahtonen Jun 25 '17 at 13:41
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I guess we are allowed to assume $z$ is real (and consequently write off the possibility that $z^2+z^{-2}=-7$). A bit unfortunate choice of notation given that $z$ often is the default complex variable. – Jyrki Lahtonen Jun 25 '17 at 13:43
7 Answers
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HINT: You could further find $$z+\frac{1}{z}=3$$ and note that $$a^3+b^3=(a+b)(a^2-ab+b^2)$$
LM2357
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Hint. Use the identity $$z^{m+n}+\frac{1}{z^{m+n}}=\left( z^m+\frac{1}{z^{m}}\right)\left(z^{n}+\dfrac{1}{z^{n}}\right)-\left(z^{m-n}+\frac{1}{z^{m-n}}\right).$$
Robert Z
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Use the same trick again to get \begin{eqnarray*} (z+\frac{1}{z})^2-2=7 \\ z+\frac{1}{z}=3 \end{eqnarray*}
\begin{eqnarray*} (z+\frac{1}{z})(z^2+\frac{1}{z^2})=(z^3+\frac{1}{z^3})+(z+\frac{1}{z})=21 \\ z^3+\frac{1}{z^3}=\color{red}{18}. \end{eqnarray*}
Donald Splutterwit
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Since $z^2+\frac{1}{z^2}=7$, we obtain $z+\frac{1}{z}=3$ or $z+\frac{1}{z}=-3$.
In the first case we obtain $$z^3+\frac{1}{z^3}=3\left(z^2+\frac{1}{z^2}-1\right)=3(7-1)=18$$
in the second case we obtain $$z^3+\frac{1}{z^3}=-3(7-1)=-18$$
Michael Rozenberg
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Note that :suppose $a=z+\frac 1z $so $$z+\frac 1z=a\\z^2+\frac{1}{z^2}=(z+\frac 1z)^2-2z\frac1z=a^2-2\\ z^4+\frac{1}{z^4}=(z^2+\frac{1}{z^2})^2-2=(a^2-2)^2-2\\ z^3+\frac{1}{z^3}=(z+\frac 1z)^3-3z\frac1z(z+\frac 1z)=a^3-3a$$ what you have now is $$(a^2-2)^2-2=47 \to\\(a^2-2)^2=49 \\\begin{cases}a^2-2=7 & a^2=9\\\to a^2-2=-7 & a^2=-5 \end{cases} \\a^2-2= 7 \\\to a^2=9 \\\to a=\pm 3 $$now
you want $$z^3+\frac{1}{z^3}=a^3-3a=\\(\pm 3)^3-3(\pm 3)$$
Khosrotash
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Try this notation for less clutter:
Let $$S_n=z^n+\frac 1{z^n}$$
It can be easily shown that
$$S_n^2=S_{2n}+2$$
Hence $$S_4=S_2^2-2=47 \qquad \Rightarrow S_2=7\\ S_2=S_1^2-2=7\qquad \Rightarrow S_1=3$$
Also, $$S_aS_b=S_{a+b}S_{a-b}$$
Putting $a=3, b=1$,
$$S_3S_1=S_4+S_2\\ S_3=\frac {S_4+S_2}{S_1}=\frac {47+7}3=\color{red}{18}$$
Hypergeometricx
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