Question

(2+I)/(3-i) (1+I) in polar form​

Answer 1

Answer: 1(cos π/2 + isin π/2)

Step - by - Step Explanation:

→ (2 + i)(1 + i)/(3 - i)

Since we are obsessed with this iota in denominator so we will rationalize it.

→ (2 + i)(1 + i)(3 + i)/(3² + 1²)

Now start opening the brackets

→ (2 + 2i + i + i²)(3 + i)/(9 + 1)

→ (1 + 3i)(3 + i)/10

→ (3 + i + 9i + 3i²)/10

→ (3 + 10i - 3)/10

→ 10i/10 → i

Polar Form

→ z = r(cos∅ + isin∅) where ∅ = arg(z) & r = |z|

→ |z| = |0 + 1i| = √(0² + 1²) = 1

→ tan∅ = 1/0 → ∅ = π/2

→ arg(z) = π/2

Thereby on substituting values we get,

→ z = 1(cos π/2 + isinπ/2) (Ans.)

Answer 2

Answer: 1(cos π/2 + isin π/2)

Step - by - Step Explanation:

→ (2 + i)(1 + i)/(3 - i)

Since we are obsessed with this iota in denominator so we will rationalize it.

→ (2 + i)(1 + i)(3 + i)/(3² + 1²)

Now start opening the brackets

→ (2 + 2i + i + i²)(3 + i)/(9 + 1)

→ (1 + 3i)(3 + i)/10

→ (3 + i + 9i + 3i²)/10

→ (3 + 10i - 3)/10

→ 10i/10 → i

Polar Form

→ z = r(cos∅ + isin∅) where ∅ = arg(z) & r = |z|

→ |z| = |0 + 1i| = √(0² + 1²) = 1

→ tan∅ = 1/0 → ∅ = π/2

→ arg(z) = π/2

Thereby on substituting values we get,

→ z = 1(cos π/2 + isinπ/2) (Ans.)