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If $\left| z \right| = \left| \omega \right|,\omega \ne 0$and$\arg \left( z \right) + \arg \left( \omega \right) = \pi $, then $z = $
$
  {\text{a}}{\text{. }} - \omega \\
  {\text{b}}{\text{. }}\omega \\
  {\text{c}}{\text{. }}\varpi \\
  {\text{d}}{\text{. }} - \varpi \\
$

Answer Verified Verified
Hint: Assume $z = \left| z \right|{e^{i\alpha }}$and $\omega = \left| \omega \right|{e^{i\beta }}$

Let, $z = \left| z \right|{e^{i\alpha }}.............\left( 1 \right),{\text{ }}\omega = \left| \omega \right|{e^{i\beta }}.........\left( 2 \right)$
Where $z$and $\omega $are complex numbers.
From equation 1,$\arg \left( z \right) = \alpha $and $\arg \left( \omega \right) = \beta $
According to question it is given that
$
  \arg \left( z \right) + \arg \left( \omega \right) = \pi \\
   \Rightarrow \alpha + \beta = \pi \\
   \Rightarrow \alpha = \pi - \beta ..........\left( 3 \right) \\
$
From equation (1) and (3)
$
  z = \left| z \right|{e^{i\alpha }} \\
   \Rightarrow z = \left| z \right|{e^{i\left( {\pi - \beta } \right)}} \\
   \Rightarrow z = \left| z \right|{e^{i\pi }}{e^{ - i\beta }}........\left( 4 \right) \\
$
Now from equation (2)
$\omega = \left| \omega \right|{e^{i\beta }}$
Now take conjugate on both sides
$
  \varpi = \overline {\left| \omega \right|{e^{i\beta }}} \\
   \Rightarrow \varpi = \left| \varpi \right|{e^{ - i\beta }} \\
   \Rightarrow {e^{ - i\beta }} = \frac{\varpi }{{\left| \varpi \right|}}..........\left( 5 \right) \\
$
Now, from equation (4) and (5)
$
   \Rightarrow z = \left| z \right|{e^{i\pi }}{e^{ - i\beta }} \\
   \Rightarrow z = \left| z \right|{e^{i\pi }}\left( {\frac{\varpi }{{\left| \varpi \right|}}} \right).......\left( 6 \right) \\
$
Now as we know modulus of any complex numbers and its conjugate both are equal so, use this property
$\left| \omega \right| = \left| \varpi \right|$
Therefore from equation (6)
$ \Rightarrow z = \left| z \right|{e^{i\pi }}\left( {\frac{\varpi }{{\left| \omega \right|}}} \right).........\left( 7 \right)$
Now it is given that
$\left| z \right| = \left| \omega \right|,\omega \ne 0$
Therefore from equation (7)
$
   \Rightarrow z = \left| z \right|{e^{i\pi }}\left( {\frac{\varpi }{{\left| z \right|}}} \right) \\
   \Rightarrow z = \varpi {e^{i\pi }}........\left( 8 \right) \\
$
Now according to Euler’s Theorem ${e^{ix}} = \cos x + i\sin x$
$ \Rightarrow {e^{i\pi }} = \cos \pi + i\sin \pi $
Now we know $\cos \pi = - 1,{\text{ }}\sin \pi = 0$
$ \Rightarrow {e^{i\pi }} = - 1 + 0 = - 1$
Therefore from equation (8)
$
   \Rightarrow z = \varpi {e^{i\pi }} \\
   \Rightarrow z = - \varpi \\
$
Hence, option (d) is correct.

Note: Whenever we face such types of problems, always assume the complex numbers in the form of $z = \left| z \right|{e^{i\alpha }}$and $\omega = \left| \omega \right|{e^{i\beta }}$, then use the given conditions to simplify it, then use the property that modulus of any complex numbers and its conjugate both are equal and finally using Euler’s Theorem we get the required result.

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