Leaving Cert Mathematics Induction (Proof): (cos θ + i sin θ)ⁿ = cos(nθ) + i

(cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ), where i² = -1 - Leaving Cert Mathematics - Question 4 - 2018

Question 4

(cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ), where i² = -1. (b) Hence, or otherwise, find (−rac{1}{2} + rac{ ext{√}3}{2} i)³ in its simplest form.

Worked Solution & Example Answer:(cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ), where i² = -1 - Leaving Cert Mathematics - Question 4 - 2018

Step 1

Prove, using induction, that if n is a positive integer then (cos θ + i sin θ)ⁿ = cos(nθ) + i sin(nθ), where i² = -1.

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Answer

Step 1: Base Case

Let's first verify the base case for n = 1.

$(cos θ + i sin θ)^1 = cos(1θ) + i sin(1θ)$ This holds true as both sides are equal.

Step 2: Inductive Step

Assume that the statement holds for n = k, i.e.,

$(cos θ + i sin θ)^k = cos(kθ) + i sin(kθ)$

We need to prove it for n = k + 1:

$(cos θ + i sin θ)^{k+1} = (cos θ + i sin θ)^k (cos θ + i sin θ).$

Using our inductive hypothesis:

$= (cos(kθ) + i sin(kθ))(cos θ + i sin θ)$

Step 3: Product Expansion

Expanding this using the distributive property:

$= cos(kθ)cos θ - sin(kθ)sin θ + i[sin(kθ)cos θ + cos(kθ)sin θ]$

Using angle addition formulas:

$= cos((k + 1)θ) + i sin((k + 1)θ).$

Thus, by induction, the proposition is true for all positive integers n.

Step 2

Hence, or otherwise, find (−1/2 + √3/2 i)³ in its simplest form.

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Answer

Step 1: Modulus and Argument

First, find the modulus and argument of the complex number:

$z = - rac{1}{2} + rac{ ext{\sqrt}3}{2} i$

The modulus is given by:

$|z| = ext{√}igg{(- rac{1}{2})^2 + igg{( rac{ ext{√}3}{2}igg{)}^2} } = 1$

Next, the argument (θ) is:

$θ = ext{arg}(z) = an^{-1}igg{( rac{ rac{ ext{√}3}{2}}{- rac{1}{2}} igg{)} = rac{2 ext{π}}{3}$

Step 2: Using De Moivre's Theorem

Now apply De Moivre's theorem for the third power:

$z^3 = |z|^3 igg{(cosigg{(3 rac{2 ext{π}}{3}igg{)} + i ext{sin}igg{(3 rac{2 ext{π}}{3}igg{)}}} = 1^3 (cos(2 ext{π}) + i sin(2 ext{π})) = 1 + 0i$

Thus, the simplest form is:

$z^3 = 1$