If omega_1 and omega_2 are two non-zero compl | vedclass

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purely imaginary number

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(D) Given $|a \omega_1 + b \omega_2| = |a \omega_1 - b \omega_2|$.Squaring both sides,we get $|a \omega_1 + b \omega_2|^2 = |a \omega_1 - b \omega_2|^2$.Using the property $|z|^2 = z \bar{z}$,we have $(a \omega_1 + b \omega_2)(a \bar{\omega}_1 + b \bar{\omega}_2) = (a \omega_1 - b \omega_2)(a \bar{\omega}_1 - b \bar{\omega}_2)$.Expanding both sides: $a^2 |\omega_1|^2 + ab \omega_1 \bar{\omega}_2 + ab \bar{\omega}_1 \omega_2 + b^2 |\omega_2|^2 = a^2 |\omega_1|^2 - ab \omega_1 \bar{\omega}_2 - ab \bar{\omega}_1 \omega_2 + b^2 |\omega_2|^2$.Canceling common terms $a^2 |\omega_1|^2$ and $b^2 |\omega_2|^2$,we get $2ab \omega_1 \bar{\omega}_2 + 2ab \bar{\omega}_1 \omega_2 = 0$.Since $a, b 
eq 0$,we have $\omega_1 \bar{\omega}_2 + \bar{\omega}_1 \omega_2 = 0$.This implies $\omega_1 \bar{\omega}_2 + \overline{\omega_1 \bar{\omega}_2} = 0$.Let $z = \frac{\omega_1}{\omega_2}$. Then $\omega_1 = z \omega_2$. Substituting this,we get $z \omega_2 \bar{\omega}_2 + \bar{z} \bar{\omega}_2 \omega_2 = 0$.Since $\omega_2 
eq 0$,$|\omega_2|^2 
eq 0$,so $z + \bar{z} = 0$.This means $2 	ext{Re}(z) = 0$,so $	ext{Re}(\frac{\omega_1}{\omega_2}) = 0$.Thus,$\frac{\omega_1}{\omega_2}$ is a purely imaginary number.

If $\omega_1$ and $\omega_2$ are two non-zero complex numbers and $a, b$ are non-zero real numbers such that $|a \omega_1 + b \omega_2| = |a \omega_1 - b \omega_2|$,then $\frac{\omega_1}{\omega_2}$ is

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