Question

# 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 \\$

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 Eulerâ€™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 Eulerâ€™s Theorem we get the required result.