For a complex number Z = a + ib,let hat{Z} = | vedclass

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(C) Given $Z = a + ib$,we have $\hat{Z} = b + ia = i(a - ib) = i\bar{Z}$.Let $Z_1 = a + ib$ and $Z_2 = c + id$.Then $Z_1 Z_2 = (ac - bd) + i(ad + bc)$

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$\frac{\hat{Z}_1 \hat{Z}_2}{i}$

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(C) Given $Z = a + ib$,we have $\hat{Z} = b + ia = i(a - ib) = i\bar{Z}$.Let $Z_1 = a + ib$ and $Z_2 = c + id$.Then $Z_1 Z_2 = (ac - bd) + i(ad + bc)$.Applying the definition of $\hat{Z}$,we get $\widehat{Z_1 Z_2} = (ad + bc) + i(ac - bd)$.Now,calculate $\hat{Z}_1 \hat{Z}_2 = (b + ia)(d + ic) = (bd + ibc + iad + i^2 ac) = (bd - ac) + i(bc + ad)$.Comparing the two,we see $\widehat{Z_1 Z_2} = -i \hat{Z}_1 \hat{Z}_2$ or similar relations depending on the definition.However,evaluating the provided options,there is a discrepancy. Given the standard form,the correct identity is $\widehat{Z_1 Z_2} = \frac{\hat{Z}_1 \hat{Z}_2}{i}$.

For a complex number $Z = a + ib$,let $\hat{Z} = b + ia$. If $Z_1$ and $Z_2$ are such complex numbers,then $\widehat{Z_1 Z_2} = $

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