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ok, so I have to calculate the conjuage of ${(8-2i)^4\over(4+3i)^5}$ using the properties such as $\overline{\left(\frac{z_1}{z_2}\right)}=\frac{\bar z_1}{\bar z_2}$ and $\overline{(z_1z_2)}=\bar z_1\bar z_2$. Using these properties, Im assuming that the answer is $\frac{(8+2i)^4}{(4-3i)^5}$ Is this correct?

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  • $\begingroup$ Yes but I was asked to not calculate this, to just use properties. Sticking to that, I'm assuming this is as far as I should go $\endgroup$
    – cele
    Oct 5 '14 at 1:27
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    $\begingroup$ That's the answer using just those two identities. $\endgroup$ Oct 5 '14 at 1:29
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$$(8-4i)(8-4i)(8-4i)(8-4i)\over(4+3i)(4+3i)(4+3i)(4+3i)(4+3i)$$ $$f(a,b) = a -bi$$ solving for the complex conjugate of the numerator $$f(8,4) = 8 - (-4)i $$ $$ f(8,4) = 8 + 4i $$ solving for the complex conjugate of the denominator $$f(4,3) = 4 - (3)i $$ $$ f(4,3) = 4 - 3i $$ thus$$\overline{(8-4i)^4\over(4+3i)^5}$$ = $$(8+4i)^4\over(4-3i)^5$$ last identity need not be used but the prove is as follows. $$\overline{z_1z_2} = \bar z_1\bar z_2$$ $$\overline{z_1z_2} = \overline{(a_1 + a_2) + (b_1 + b_2)i} $$ $$(a_1 + a_2) - (b_1 + b_2)i$$ $$(a_1 - b_1i)(a_2 - b_2i)$$

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