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Question: If \(\left| z \right| = \left| \omega \right|,\omega \ne 0\)and\(\arg \left( z \right) + \arg \left(...

If ∣z∣=∣ω∣,ω≠0\left| z \right| = \left| \omega \right|,\omega \ne 0andarg⁡(z)+arg⁡(ω)=π\arg \left( z \right) + \arg \left( \omega \right) = \pi , then z=z =
a. −ω b. ω c. ϖ d. −ϖ  {\text{a}}{\text{. }} - \omega \\\ {\text{b}}{\text{. }}\omega \\\ {\text{c}}{\text{. }}\varpi \\\ {\text{d}}{\text{. }} - \varpi \\\

Explanation

Solution

Hint: Assume z=∣z∣eiαz = \left| z \right|{e^{i\alpha }}and ω=∣ω∣eiβ\omega = \left| \omega \right|{e^{i\beta }}

Let, z=∣z∣eiα.............(1), ω=∣ω∣eiβ.........(2)z = \left| z \right|{e^{i\alpha }}.............\left( 1 \right),{\text{ }}\omega = \left| \omega \right|{e^{i\beta }}.........\left( 2 \right)
Where zzand ω\omega are complex numbers.
From equation 1,arg⁡(z)=α\arg \left( z \right) = \alpha and arg⁡(ω)=β\arg \left( \omega \right) = \beta
According to question it is given that
arg⁡(z)+arg⁡(ω)=π ⇒α+β=π ⇒α=π−β..........(3)  \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=∣z∣eiα ⇒z=∣z∣ei(π−β) ⇒z=∣z∣eiπe−iβ........(4)  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)
ω=∣ω∣eiβ\omega = \left| \omega \right|{e^{i\beta }}
Now take conjugate on both sides
ϖ=∣ω∣eiβ‾ ⇒ϖ=∣ϖ∣e−iβ ⇒e−iβ=ϖ∣ϖ∣..........(5)  \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)
⇒z=∣z∣eiπe−iβ ⇒z=∣z∣eiπ(ϖ∣ϖ∣).......(6)  \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
∣ω∣=∣ϖ∣\left| \omega \right| = \left| \varpi \right|
Therefore from equation (6)
⇒z=∣z∣eiπ(ϖ∣ω∣).........(7)\Rightarrow z = \left| z \right|{e^{i\pi }}\left( {\frac{\varpi }{{\left| \omega \right|}}} \right).........\left( 7 \right)
Now it is given that
∣z∣=∣ω∣,ω≠0\left| z \right| = \left| \omega \right|,\omega \ne 0
Therefore from equation (7)
⇒z=∣z∣eiπ(ϖ∣z∣) ⇒z=ϖeiπ........(8)  \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 eix=cos⁡x+isin⁡x{e^{ix}} = \cos x + i\sin x
⇒eiπ=cos⁡π+isin⁡π\Rightarrow {e^{i\pi }} = \cos \pi + i\sin \pi
Now we know cos⁡π=−1, sin⁡π=0\cos \pi = - 1,{\text{ }}\sin \pi = 0
⇒eiπ=−1+0=−1\Rightarrow {e^{i\pi }} = - 1 + 0 = - 1
Therefore from equation (8)
⇒z=ϖeiπ ⇒z=−ϖ  \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=∣z∣eiαz = \left| z \right|{e^{i\alpha }}and ω=∣ω∣eiβ\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.