Question 1209757: If z is a complex number satisfying
z + \frac{1}{z} = \sqrt{2},
calculate
z^{10} + \frac{1}{z^{10}}.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let's solve this problem step-by-step.
**1. Rewrite the Given Equation:**
We are given:
z + 1/z = √2
Multiply both sides by z:
z² + 1 = √2z
Rearrange to form a quadratic equation:
z² - √2z + 1 = 0
**2. Solve for z:**
Using the quadratic formula:
z = [√2 ± √((√2)² - 4(1)(1))] / 2
z = [√2 ± √(2 - 4)] / 2
z = [√2 ± √(-2)] / 2
z = [√2 ± i√2] / 2
z = (√2 / 2) ± i(√2 / 2)
z = (1/√2) ± i(1/√2)
**3. Express z in Polar Form:**
We can express z in polar form as z = r(cos θ + i sin θ), where r is the magnitude and θ is the argument.
* Magnitude (r):
r = √[(1/√2)² + (1/√2)²] = √(1/2 + 1/2) = √1 = 1
* Argument (θ):
Since cos θ = 1/√2 and sin θ = ±1/√2, we have:
* If z = (1/√2) + i(1/√2), then θ = π/4.
* If z = (1/√2) - i(1/√2), then θ = -π/4.
Thus, we have:
* z = cos(π/4) + i sin(π/4) or
* z = cos(-π/4) + i sin(-π/4)
**4. Use De Moivre's Theorem:**
De Moivre's Theorem states that for any complex number z = r(cos θ + i sin θ) and integer n:
z^n = r^n (cos(nθ) + i sin(nθ))
In our case, r = 1, so:
z^n = cos(nθ) + i sin(nθ)
**5. Calculate z^10:**
* If θ = π/4:
z^10 = cos(10π/4) + i sin(10π/4) = cos(5π/2) + i sin(5π/2) = cos(π/2) + i sin(π/2) = 0 + i(1) = i
* If θ = -π/4:
z^10 = cos(-10π/4) + i sin(-10π/4) = cos(-5π/2) + i sin(-5π/2) = cos(-π/2) + i sin(-π/2) = 0 + i(-1) = -i
**6. Calculate 1/z^10:**
* If z^10 = i, then 1/z^10 = 1/i = -i.
* If z^10 = -i, then 1/z^10 = 1/(-i) = i.
**7. Calculate z^10 + 1/z^10:**
* If z^10 = i and 1/z^10 = -i, then z^10 + 1/z^10 = i + (-i) = 0.
* If z^10 = -i and 1/z^10 = i, then z^10 + 1/z^10 = -i + i = 0.
**Final Answer:**
z^10 + 1/z^10 = 0
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