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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(iii) Complex Numbers – Exercise 13.4 – Q.1(iv) Polar Form, z = r (cos ⁡θ + i sin ⁡θ) Complex Numbers – Exercise 13.4 – Q.1(v) 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)3 = (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)).

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