Let r and respectively be the modulus and amplitude of the complex number , then is equal to [2024]

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Q.
Let $r$ and $θ$ respectively be the modulus and amplitude of the complex number $z = 2 - i (2 \tan \frac{5 π}{8}),$ then $(r, θ)$ is equal to [2024]

1 $(2 \sec \frac{3 π}{8}, \frac{5 π}{8})$

2 $(2 \sec \frac{3 π}{8}, \frac{3 π}{8})$

3 $(2 \sec \frac{11 π}{8}, \frac{11 π}{8})$

4 $(2 \sec \frac{5 π}{8}, \frac{3 π}{8})$

Ans.
(2)

We have, $z = 2 - i (2 \tan \frac{5 π}{8})$

$= 2 + 2 i \tan (π - \frac{5 π}{8}) \Rightarrow z = 2 + 2 i \tan \frac{3 π}{8}$

$\Rightarrow z = 2 (\cos \frac{3 π}{8} + i \sin \frac{3 π}{8}) \sec \frac{3 π}{8}$

So, $r = 2 \sec \frac{3 π}{8}, θ = \frac{3 π}{8}$

Solution

Q.
Let $r$ and $θ$ respectively be the modulus and amplitude of the complex number $z = 2 - i (2 \tan \frac{5 π}{8}),$ then $(r, θ)$ is equal to [2024]

1 $(2 \sec \frac{3 π}{8}, \frac{5 π}{8})$

2 $(2 \sec \frac{3 π}{8}, \frac{3 π}{8})$

3 $(2 \sec \frac{11 π}{8}, \frac{11 π}{8})$

4 $(2 \sec \frac{5 π}{8}, \frac{3 π}{8})$

Ans.
(2)

We have, $z = 2 - i (2 \tan \frac{5 π}{8})$

$= 2 + 2 i \tan (π - \frac{5 π}{8}) \Rightarrow z = 2 + 2 i \tan \frac{3 π}{8}$

$\Rightarrow z = 2 (\cos \frac{3 π}{8} + i \sin \frac{3 π}{8}) \sec \frac{3 π}{8}$

So, $r = 2 \sec \frac{3 π}{8}, θ = \frac{3 π}{8}$

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Solution

Q. Let r and θ respectively be the modulus and amplitude of the complex number z=2-i(2tan5π8), then (r,θ) is equal to [2024]

1 (2sec3π8,5π8)

2 (2sec3π8,3π8)

3 (2sec11π8,11π8)

4 (2sec5π8,3π8)

Ans. (2) We have, z=2-i(2tan5π8) =2+2itan(π-5π8)⇒z=2+2itan3π8 ⇒z=2(cos3π8+isin3π8)sec3π8 So, r=2sec3π8, θ=3π8

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Q.
Let $r$ and $θ$ respectively be the modulus and amplitude of the complex number $z = 2 - i (2 \tan \frac{5 π}{8}),$ then $(r, θ)$ is equal to [2024]

1 $(2 \sec \frac{3 π}{8}, \frac{5 π}{8})$

2 $(2 \sec \frac{3 π}{8}, \frac{3 π}{8})$

3 $(2 \sec \frac{11 π}{8}, \frac{11 π}{8})$

4 $(2 \sec \frac{5 π}{8}, \frac{3 π}{8})$

Ans.
(2)

We have, $z = 2 - i (2 \tan \frac{5 π}{8})$

$= 2 + 2 i \tan (π - \frac{5 π}{8}) \Rightarrow z = 2 + 2 i \tan \frac{3 π}{8}$

$\Rightarrow z = 2 (\cos \frac{3 π}{8} + i \sin \frac{3 π}{8}) \sec \frac{3 π}{8}$

So, $r = 2 \sec \frac{3 π}{8}, θ = \frac{3 π}{8}$