Write 1+iroot 3 and 1-iroot3 in trig form and simplify (1+iroot3)^9 +(1-iroot3)^9 in x+iy form.
1+√3 i = 2cis π/3
1-√3 i = 2cis -π/3
so,
(1+√3 i)^9 = 2^9 cis (π/3 * 9) = 512 cis 3π = -512
now finish it off
1+iroot 3
= 1 + √3
tanθ = √3/1 -----> θ = 60 degrees or θ = π/3 radians
r = √(1 + 3) = 2
1 + √3 = 2(cosπ/3 + i sinπ/3)
so (1 + √3)^9 = 2^9(cos 9*π/3 + i sin 9*π/3)
= 512(cos 3π + i sin 3π)
= 512(-1 + 0i) = -512
do the same for 1 - √3 and then (1 - √3)^9
To write 1 + i√3 in trigonometric form, we can convert it to exponential form and then convert to trigonometric form using Euler's formula:
1 + i√3 = √(1^2 + (√3)^2) * (cosθ + isinθ)
where θ is the argument of 1 + i√3.
Using the equation:
√(1^2 + (√3)^2) = 2
θ = arctan(√3/1) = π/3
Therefore, we have:
1 + i√3 = 2 * (cos(π/3) + isin(π/3))
Simplifying, we have:
1 + i√3 = 2(cos(π/3) + i * sin(π/3))
Next, to express (1 + i√3)^9 in x + iy form, we can use De Moivre's formula:
(z^m) = r^m (cos(mθ) + isin(mθ))
where z = 1 + i√3, r = 2 (from above), and m = 9.
Using De Moivre's formula, we have:
(1 + i√3)^9 = (2)^9 (cos(9π/3) + i * sin(9π/3))
Simplifying further:
(1 + i√3)^9 = (2)^9 (cos(3π) + i * sin(3π))
Since cos(3π) = 1 and sin(3π) = 0, we have:
(1 + i√3)^9 = 2^9 (1 + i * 0)
Finally, simplifying:
(1 + i√3)^9 = 512 + 0i
Therefore, (1 + i√3)^9 in x + iy form is 512 + 0i.
To express a complex number in trigonometric or polar form, we need to rewrite it in terms of its magnitude and argument.
Let's begin with 1 + i√3. To find its magnitude, we can use the Pythagorean theorem:
|1 + i√3| = √(1² + (√3)²) = √(1 + 3) = 2.
To find its argument, we can use the inverse tangent function:
θ = arctan(√3 / 1) = arctan(√3) ≈ π/3.
Therefore, 1 + i√3 in trigonometric or polar form is 2 ∠ π/3.
Now, let's express 1 - i√3 in trigonometric form. Following the same steps:
|1 - i√3| = √(1² + (-√3)²) = √(1 + 3) = 2.
θ = arctan(-√3 / 1) = arctan(-√3) ≈ -π/3.
So, 1 - i√3 in trigonometric form is 2 ∠ -π/3.
Next, let's simplify (1 + i√3)^9 + (1 - i√3)^9 in the x + iy form.
We can use De Moivre's theorem, which states that (r ∠ θ)^n = r^n ∠ nθ.
Simplifying (1 + i√3)^9 gives us:
(1 + i√3)^9 = 2^9 ∠ (9 * π/3) = 512 ∠ 3π.
Simplifying (1 - i√3)^9 gives us:
(1 - i√3)^9 = 2^9 ∠ (9 * -π/3) = 512 ∠ -3π.
Adding these two expressions, we get:
(1 + i√3)^9 + (1 - i√3)^9 = 512 ∠ 3π + 512 ∠ -3π.
To simplify this expression, we can combine the angles:
= 512 ∠ 3π - 512 ∠ 3π.
This results in the expression equaling zero:
= 0.
Therefore, (1 + i√3)^9 + (1 - i√3)^9 simplifies to 0 in the x + iy form.