Find the modulus and the principal argument of the complex number $\frac{i - 1}{i (1 - c o s \frac{2 π}{5}) + s i n \frac{2 π}{5}}$

\frac{c o s e c (\frac{π}{5})}{\sqrt{2}}

\frac{9 π}{20}

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\frac{s i n (\frac{π}{5})}{\sqrt{2}}

\frac{11 π}{20}

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\frac{c o s e c (\frac{π}{5})}{\sqrt{2}}

\frac{11 π}{20}

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\frac{s i n (\frac{π}{5})}{\sqrt{2}}

\frac{9 π}{20}

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Solution

The correct option is C $\frac{c o s e c (\frac{π}{5})}{\sqrt{2}}$ , $\frac{11 π}{20}$
$z = \frac{i - 1}{i (1 - c o s \frac{2 π}{5}) + s i n \frac{2 π}{5}}$
$= \frac{i - 1}{i 2 s i n^{2} \frac{π}{5} + 2 s i n \frac{π}{5} c o s \frac{π}{5}}$
$= \frac{i - 1}{(2 s i n \frac{π}{5}) (c o s \frac{π}{5} + i s i n \frac{π}{5})}$
$| z | = \frac{| i - 1 |}{(2 s i n \frac{π}{5}) ∣ ∣ (c o s \frac{π}{5} + i s i n \frac{π}{5}) ∣ ∣}$
$= \frac{\sqrt{2}}{(2 s i n \frac{π}{5}) ∣ ∣ (c o s \frac{π}{5} + i s i n \frac{π}{5}) ∣ ∣}$
$= \frac{1}{\sqrt{2}} c o s e c \frac{π}{5}$
$a r g z = a r g ⎡ ⎢ ⎢ ⎣ \frac{i - 1}{(2 s i n \frac{π}{5}) (c o s \frac{π}{5} + i s i n \frac{π}{5})} ⎤ ⎥ ⎥ ⎦$
$= a r g (- 1 + i) - a r g (2 s i n \frac{π}{5}) - a r g (c o s \frac{π}{5} + i s i n \frac{π}{5})$
$= \frac{3 π}{4} - 0 - \frac{π}{5} = \frac{11 π}{20}$
Ans: B

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