Question

Let $z = \frac{1 + 2 (\sin θ) i}{1 - i (\sin θ)}$

(i) Find the number of values of $θ \in [0, 4 π]$ such that $z$ is purely imaginary.

(ii) Find the sum of all values of $θ \in [0, 4 π]$ such that $z$ is purely real.

Accepted Answer

We have,

$z = \frac{1 + 2 (\sin θ) i}{1 - i (\sin θ)}$

$\Rightarrow z = \frac{1 + 2 (\sin θ) i}{1 - i (\sin θ)} \times \frac{1 + i (\sin θ)}{1 + i (\sin θ)}$

$\Rightarrow z = \frac{1 + 2 (\sin θ) i + i (\sin θ) - 2 (\sin^{2} θ)}{1 - i^{2} (\sin^{2} θ)}$

$\Rightarrow z = \frac{1 - 2 (\sin^{2} θ) + 3 (\sin θ) i}{1 + (\sin^{2} θ)}$

$\Rightarrow z = \{\frac{1 - 2 (\sin^{2} θ)}{1 + (\sin^{2} θ)}\} + i \{\frac{3 (\sin θ)}{1 + (\sin^{2} θ)}\}$

$(i)$ For purely imaginary number, real part must be zero, hence

$\frac{1 - 2 (\sin^{2} θ)}{1 + (\sin^{2} θ)} = 0$

$\Rightarrow 1 - 2 (\sin^{2} θ) = 0$

$\Rightarrow \sin θ = \pm \frac{1}{\sqrt{2}} = \sin (\frac{π}{4})$

For $θ \in [0, 4 π]$ , we have $θ = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}, \frac{9 π}{4}, \frac{11 π}{4}, \frac{13 π}{4}, \frac{15 π}{4}, \frac{17 π}{4}$

Number of solutions is $8$ .

$(ii)$ For purely real number, imaginary part must be zero, hence

$\frac{3 (\sin θ)}{1 + (\sin^{2} θ)} = 0$

$\Rightarrow \sin θ = 0$

$\Rightarrow θ = π, 2 π, 3 π, 4 π$

Sum of solutions is $= π + 2 π + 3 π + 4 π = 10 π$

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