Question

complex numbers......​

Answer 1

To prove :

$\star \: \frac{1}{1 + 2 \omega} + \frac{1}{2 +\omega } - \frac{1}{1 +\omega } = 0$

Solution:

• As we know that :

$\star \: 1 + \omega + {\omega}^{2} = 0 \\ \\ \star \: 1 + \omega = - {\omega}^{2} \\ \\ \star \: 1 + 2\omega = \omega - {\omega}^{2} \\ \\ \star \: 2 + \omega = 1 - {\omega}^{2}$

Now LHS :

$\implies \: \frac{1}{1 + 2\omega} + \frac{1}{2 +\omega } - \frac{1}{1 + \omega} \\ \\ \\ \implies \: \frac{1}{\omega(1 - \omega)} + \frac{1}{(1 - \omega)(1 +\omega) } + \frac{1}{ {\omega}^{2} } \\ \\ \implies \: \frac{\omega(1 +\omega) + {\omega}^{2} + 1 - {\omega}^{2} }{ {\omega}^{2}(1 - {\omega}^{2} )} \\ \\ \implies \: \frac{1 +\omega + {\omega}^{2} }{ {\omega}^{2} (1 - {\omega}^{2} )} \\ \\ \implies \: \frac{0}{ {\omega}^{2} (1 - {\omega}^{2} )} \\ \\ \implies \: 0$

LHS= RHS

$\huge \red{ \mathfrak{proved}}$

Answer 2

Answer:

ANSWER IS 0

Step-by-step explanation:

HERE IS U R ANSWER DEAR PLZ FOLLW ME