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# Intermediate Algebra

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Let $$z$$ be a complex number such that

$$z+\frac{1}{z}=1$$

What is $$z^3$$?

Nov 20, 2023

#1
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Squaring the equation z + 1/z = 1 gives z^2 + 2 + 1/z^2 = 1. Then multiplying by z^2 gives z^4 + 2z^2 + 1 = z^2. Rearranging, we get z^4 + 2z^2 + 1 - z^2 = 0, which factors as (z^2 + 1)(z^2 + 1) = 0. Since z^2 + 1 = 0 has no solutions, z^2 = -1. Taking the square root of both sides, we find z = i or z = -i.

If z = i, then z^3 = i^3 = -i.

Alternatively, if z = -i, then z^3 is also equal to -i.

Therefore,  in either case, z^3 = -i.

Nov 20, 2023
#3
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Yeah, I managed to solve it 30 minutes later. (Sorry, the correct answer was -1.)

Anyways, thanks for the answer. What I did was take

$$z + \frac{1}{z}=1$$ and multiply it by $$z^2$$, getting $$z^3-z^2+z=0$$. However, when you multiply $$z+ \frac{1}{z}=1$$ by $$z$$, you will get

$$z^2-z=1$$

From which you can replace the $$-z^2+z$$ with $$-1$$ (The negative of the equation above) to get

$$z^3-1=0$$

From which you can easily get $$z^3=-1$$

Nov 21, 2023
#4
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Sorry I meant $$z^3+1=0$$, and $$z^2-z=-1$$.

Sorry if I confused anyone.

JovenlyCosmo  Nov 22, 2023
edited by JovenlyCosmo  Nov 22, 2023