%5E6-in-polar-form.jpeg "(1+i)^6 In Polar Form")
Finding the Polar Form of (1+i)^6
This article will guide you through finding the polar form of the complex number (1 + i)^6.
1. Understanding Polar Form
Before we proceed, let's recall what polar form is:
A complex number z can be represented in polar form as:
z = r(cos θ + i sin θ)
Where:
- r is the magnitude or modulus of z, found by √(x² + y²)
- θ is the angle or argument of z, found using arctan(y/x)
2. Finding the Magnitude and Angle of (1+i)
- Magnitude:
- |1+i| = √(1² + 1²) = √2
- Angle:
- θ = arctan(1/1) = π/4
Therefore, (1+i) in polar form is √2 (cos(π/4) + i sin(π/4))
3. Using De Moivre's Theorem
De Moivre's Theorem states that for any complex number in polar form and any integer n:
(r(cos θ + i sin θ))ⁿ = rⁿ (cos(nθ) + i sin(nθ))
Applying this to our problem:
(1+i)⁶ = (√2 (cos(π/4) + i sin(π/4)))⁶
(1+i)⁶ = (√2)⁶ (cos(6π/4) + i sin(6π/4))
(1+i)⁶ = 8 (cos(3π/2) + i sin(3π/2))
4. Simplifying the Result
Since cos(3π/2) = 0 and sin(3π/2) = -1, the final polar form of (1+i)⁶ is:
**(1+i)⁶ = 8(0 - i) = ** -8i
Conclusion
Therefore, (1+i)⁶ in polar form is -8i, or equivalently 8 (cos(3π/2) + i sin(3π/2)). By utilizing De Moivre's Theorem, we were able to efficiently find the polar form of this complex number raised to a power.