Question -
Answer -
(i) 1 + i tan α
Given: Z = 1 + i tan α
We know that the polarform of a complex number Z = x + iy is given by Z = |Z| (cos θ + i sin θ)
Where,
|Z| = modulus of complexnumber = √(x2 +y2)
θ = arg (z) = argumentof complex number = tan-1 (|y| / |x|)
We also know that tanα is a periodic function with period π.
So α is lying in theinterval [0, π/2) ∪ (π/2, π].
Let us consider case1:
α ∈ [0, π/2)
So now,
|Z| = r = √(x2 + y2)
= √(12 + tan2 α)
= √( sec2 α)
= |sec α| since, sec αis positive in the interval [0, π/2)
θ = tan-1 (|y|/ |x|)
= tan-1 (tanα / 1)
= tan-1 (tanα)
= α since, tan α ispositive in the interval [0, π/2)
∴ Polar form is Z = secα (cos α + i sin α)
Let us consider case2:
α ∈ (π/2, π]
So now,
|Z| = r = √(x2 + y2)
= √(12 + tan2 α)
= √( sec2 α)
= |sec α|
= – sec α since, sec αis negative in the interval (π/2, π]
θ = tan-1 (|y|/ |x|)
= tan-1 (tanα / 1)
= tan-1 (tanα)
= -π + α since, tan αis negative in the interval (π/2, π]
θ = -π + α [since, θlies in 4th quadrant]
Z = -sec α (cos (α –π) + i sin (α – π))
∴ Polar form is Z =-sec α (cos (α – π) + i sin (α – π))
(ii) tan α – i
Given: Z = tan α – i
We know that the polarform of a complex number Z = x + iy is given by Z = |Z| (cos θ + i sin θ)
Where,
|Z| = modulus ofcomplex number = √(x2 +y2)
θ = arg (z) = argumentof complex number = tan-1 (|y| / |x|)
We also know that tanα is a periodic function with period π.
So α is lying in theinterval [0, π/2) ∪ (π/2, π].
Let us consider case1:
α ∈ [0, π/2)
So now,
|Z| = r = √(x2 + y2)
= √(tan2 α + 12)
= √( sec2 α)
= |sec α| since, sec αis positive in the interval [0, π/2)
= sec α
θ = tan-1 (|y|/ |x|)
= tan-1 (1/tanα)
= tan-1 (cotα) since, cot α is positive in the interval [0, π/2)
= α – π/2 [since, θlies in 4th quadrant]
Z = sec α (cos (α –π/2) + i sin (α – π/2))
∴ Polar form is Z = secα (cos (α – π/2) + i sin (α – π/2))
Let us consider case2:
α ∈ (π/2, π]
So now,
|Z| = r = √(x2 + y2)
= √(tan2 α + 12)
= √( sec2 α)
= |sec α|
= – sec α since, sec αis negative in the interval (π/2, π]
θ = tan-1 (|y|/ |x|)
= tan-1 (1/tanα)
= tan-1 (cotα)
= π/2 + α since, cot αis negative in the interval (π/2, π]
θ = π/2 + α [since, θlies in 3th quadrant]
Z = -sec α (cos (π/2 +α) + i sin (π/2 + α))
∴ Polar form is Z =-sec α (cos (π/2 + α) + i sin (π/2 + α))
(iii) 1 – sin α + icos α
Given: Z = 1 – sin α +i cos α
By using the formulas,
Sin2 θ+ cos2 θ = 1
Sin 2θ = 2 sin θ cos θ
Cos 2θ = cos2 θ– sin2 θ
So,
z= (sin2(α/2)+ cos2(α/2) – 2 sin(α/2) cos(α/2)) + i (cos2(α/2)– sin2(α/2))
= (cos(α/2) –sin(α/2))2 + i (cos2(α/2) – sin2(α/2))
We know that the polarform of a complex number Z = x + iy is given by Z = |Z| (cos θ + i sin θ)
Where,
|Z| = modulus ofcomplex number = √(x2 +y2)
θ = arg (z) = argumentof complex number = tan-1 (|y| / |x|)
Now,
We know that sine andcosine functions are periodic with period 2π
Here we have 3intervals:
0 ≤ α ≤ π/2
π/2 ≤ α ≤ 3π/2
3π/2 ≤ α ≤ 2π
Let us consider case1:
In the interval 0 ≤ α≤ π/2
Cos (α/2) > sin(α/2) and also 0 < π/4 + α/2 < π/2
So,
∴ Polar form is Z= √2 (cos (α/2) – sin(α/2)) (cos (π/4 + α/2) + i sin (π/4 + α/2))
Let us consider case2:
In the interval π/2 ≤α ≤ 3π/2
Cos (α/2) < sin(α/2) and also π/2 < π/4 + α/2 < π
So,
Since, (1 – sin α)> 0 and cos α < 0 [Z lies in 4th quadrant]
= α/2 – 3π/4
∴ Polar form is Z = –√2 (cos (α/2) – sin (α/2)) (cos (α/2 –3π/4) + i sin (α/2 – 3π/4))
Let us consider case3:
In the interval 3π/2 ≤α ≤ 2π
Cos (α/2) < sin(α/2) and also π < π/4 + α/2 < 5π/4
So,
θ = tan-1 (tan(π/4 + α/2))
= π – (π/4 + α/2)[since, θ lies in 1st quadrant and tan’s period is π]
= α/2 – 3π/4
∴ Polar form is Z = –√2 (cos (α/2) – sin (α/2)) (cos (α/2 –3π/4) + i sin (α/2 – 3π/4))
(iv) (1 – i) / (cos π/3 + isin π/3)
Given: Z = (1 – i) /(cos π/3 + i sin π/3)
Let us multiply anddivide by (1 – i√3), we get
We know that the polarform of a complex number Z = x + iy is given by Z = |Z| (cos θ + i sin θ)
Where,
|Z| = modulus ofcomplex number = √(x2 +y2)
θ = arg (z) = argumentof complex number = tan-1 (|y| / |x|)
Now,
Since x < 0, y <0 complex number lies in 3rd quadrant and the value of θ is 1800≤θ≤-900.
= tan-1 (2+ √3)
= -7π/12
Z = √2 (cos (-7π/12) + i sin (-7π/12))
= √2 (cos (7π/12) – i sin (7π/12))
∴ Polar form of (1 – i)/ (cos π/3 + i sin π/3) is √2(cos (7π/12) – i sin (7π/12))