Let z - cos- - isin-cos- - isin- -4 - - - -2- then z will be

We have, z = nbsp; (cosθ + isinθcosθ isinθ) (z)=(cosθ + isinθ) (cosθ isinθ)+2kπ,k∈ℤ =θ ( θ)+2kπ =2θ+2kπ Hence for k=0 (z)=2θ ,π2 lt; 2θ lt; π ...

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Standard XII

Mathematics

Geometrical Representation of Argument and Modulus

Let z = cosθ ...

Question

Let $z = (\frac{cos θ + i sin θ}{cos θ - i sin θ}), \frac{π}{4} < θ < \frac{π}{2},$ then $arg (z)$ will be

2 θ

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π - 2 θ

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θ

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π + 2 θ

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Solution

The correct option is A $2 θ$
We have,
$z = (\frac{cos θ + i sin θ}{cos θ - i sin θ})$
$arg (z) = arg (cos θ + i sin θ) - arg (cos θ - i sin θ) + 2 k π, k \in Z = θ - (- θ) + 2 k π = 2 θ + 2 k π$
Hence for $k = 0$
$arg (z) = 2 θ, \frac{π}{2} < 2 θ < π$

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Geometrical Representation of Argument and Modulus

Standard XII Mathematics

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Let z=(cosθ+isinθcosθ−isinθ),π4<θ<π2, then arg(z) will be

The correct option is A 2θWe have, z=(cosθ+isinθcosθ−isinθ) arg(z)=arg(cosθ+isinθ)−arg(cosθ−isinθ)+2kπ,k∈Z=θ−(−θ)+2kπ=2θ+2kπ Hence for k=0 arg(z)=2θ,π2<2θ<π

Let $z = (\frac{cos θ + i sin θ}{cos θ - i sin θ}), \frac{π}{4} < θ < \frac{π}{2},$ then $arg (z)$ will be

The correct option is A $2 θ$
We have,
$z = (\frac{cos θ + i sin θ}{cos θ - i sin θ})$
$arg (z) = arg (cos θ + i sin θ) - arg (cos θ - i sin θ) + 2 k π, k \in Z = θ - (- θ) + 2 k π = 2 θ + 2 k π$
Hence for $k = 0$
$arg (z) = 2 θ, \frac{π}{2} < 2 θ < π$