If tan- -i- where -2-2can be written as r ei - then r- - is

Given: z=r e^i β=tanθ i,θ∈( π2,π2) ⇒ nbsp;|z|=√(1+tan^2θ) ⇒ nbsp;|z|=√(^2θ)=|θ|=θ Now, Case 1: If θ∈(0,π2), then Re(z) gt;0 and z lies in 4^th quadra ...

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Byju's Answer

Standard XIII

Mathematics

Euler's Representation

If tanθ -i, w...

Question

If $tan θ - i,$ where $θ \in (- \frac{π}{2}, \frac{π}{2})$
can be written as $r e^{i β}$ , then $(r, β)$ is

(sec θ, θ - \frac{π}{2})

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(sec θ, θ + \frac{π}{2})

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(cosec θ, θ - \frac{π}{2})

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(cosec θ, \frac{π}{2} - θ)

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Solution

The correct option is A $(sec θ, θ - \frac{π}{2})$
$z = r e^{i β} = tan θ - i, θ \in (- \frac{π}{2}, \frac{π}{2}) \Rightarrow | z | = \sqrt{1 + {tan}^{2} θ} \Rightarrow | z | = \sqrt{{sec}^{2} θ} = | sec θ | = sec θ$

Now,
Case $1 :$ If $θ \in (0, \frac{π}{2})$ , then
$R e (z) > 0$ and $z$ lies in $4^{t h}$ quadrant.
$tan β = ∣ ∣ ∣ \frac{- 1}{tan θ} ∣ ∣ ∣ = cot θ = tan (\frac{π}{2} - θ) \Rightarrow β = \frac{π}{2} - θ ∴ arg (z) = - β = θ - \frac{π}{2}$

Case $2 :$ If $θ \in (- \frac{π}{2}, 0)$ , then
$R e (z) < 0$ and $z$ lies in $3^{r d}$ quadrant.
$tan β = ∣ ∣ ∣ \frac{- 1}{tan θ} ∣ ∣ ∣ = \frac{1}{- tan θ} = - tan (\frac{π}{2} - θ) = tan (θ - \frac{π}{2}) \Rightarrow β = θ - \frac{π}{2} ∴ arg (z) = - π + β = - \frac{π}{2} + θ$

Hence, $(r, β) = (sec θ, θ - \frac{π}{2})$

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Q. If

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Euler's Representation

Standard XIII Mathematics

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If tanθ−i, where θ∈(−π2,π2) can be written as reiβ, then (r,β) is

If $tan θ - i,$ where $θ \in (- \frac{π}{2}, \frac{π}{2})$
can be written as $r e^{i β}$ , then $(r, β)$ is