Question 1209757
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