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Chapter 13: Complex Numbers – Exercise 13.4

Complex Numbers – Exercise 13.4 – Q.1(i)

The polar form of a complex number z = x + iy, is given by z = |z| (cos⁡ θ + i sin ⁡θ)

where,

Let z = 1 + i

∵ x, y > 0, so θ lies in first quadrant

Now,

Complex Numbers – Exercise 13.4 – Q.1(ii)

The polar form of a complex number z = x + iy, is givenby z = |z| (cos ⁡θ + i sin ⁡θ)

Where,

= 2

∵ x = √3. 0 & y = 1 > 0,

∴ θ  lies in first quadrant

Hence

= tan-1⁡  (∵ tan-1)⁡ (tan⁡ x) = x)

polar form is given by z = |z|(cos ⁡θ + i sin ⁡θ)

Complex Numbers – Exercise 13.4 – Q.1(iv)

Polar Form, z = r (cos ⁡θ + i sin ⁡θ)

Complex Numbers – Exercise 13.4 – Q.1(vi)

The polar form of a  complex number z = x + iy, is given by z = |z| (cos ⁡θ + i sin ⁡θ)

where,

Complex Numbers – Exercise 13.4 – Q.1(vii)

The polar form of a complex number z = x + iy, is given by z = |z| (cos ⁡θ + i sin ⁡θ)

Where,

Here z is  already inpolar form

Complex Numbers – Exercise 13.4 – Q.1(viii)

The polar form of a complex number z = x + iy, is given by z = |z| (cos ⁡θ + i sin ⁡θ)

Where,

= 8

Here x = – 4 < 0 & y = 4R3 > 0, ∴ θ lies in quadrant II

Complex Numbers – Exercise 13.4 – Q.2

z = (i25)= (i)3 = – i

|z| = 1,

Complex Numbers – Exercise 13.4 – Q.3(i)

Let z = 1 + i tan⁡ α

tan⁡ α is periodic fun function with period π.

tan ⁡β = |tan ⁡α| = tan ⁡α

⟹ β = α

As z is represented by a point in first quadrant.

∴ arg⁡(z) = β = α

So polar form of z is sec⁡α (cos ⁡α + i sin ⁡α)

tan⁡ β = |tan ⁡α| = – tan⁡ α = tan⁡(π – α)

⟹ β = π – α

As z is represented by a point in fourth quadrant.

∴ arg⁡(z) = – β = α – π.

So polar form of z is-sec⁡α (cos⁡ (α – π) + i sin⁡ (α – π)).