The multiplicative inverse of the complex num | vedclass

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Question

(B) Let the complex number be $z = \sin 	heta + i \cos 	heta$. The multiplicative inverse of $z$ is given by $\frac{1}{z}$. $\frac{1}{z} = \frac{1}{

Vedclass · Accepted Answer

$(\sin 	heta, -\cos 	heta)$

Vedclass · Answer

(B) Let the complex number be $z = \sin 	heta + i \cos 	heta$. The multiplicative inverse of $z$ is given by $\frac{1}{z}$. $\frac{1}{z} = \frac{1}{\sin 	heta + i \cos 	heta}$. To simplify,multiply the numerator and denominator by the conjugate $(\sin 	heta - i \cos 	heta)$: $\frac{1}{z} = \frac{\sin 	heta - i \cos 	heta}{(\sin 	heta + i \cos 	heta)(\sin 	heta - i \cos 	heta)}$. Using the identity $(a+ib)(a-ib) = a^2 + b^2$: $\frac{1}{z} = \frac{\sin 	heta - i \cos 	heta}{\sin^2 	heta + \cos^2 	heta}$. Since $\sin^2 	heta + \cos^2 	heta = 1$,we get: $\frac{1}{z} = \sin 	heta - i \cos 	heta$. Thus,the multiplicative inverse is $(\sin 	heta, -\cos 	heta)$.

The multiplicative inverse of the complex number $(\sin \theta, \cos \theta)$ is

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