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Solve: $\large\frac{(2+3i)(2-3i)}{1-i}=$

$\begin{array}{1 1} (a)\; \frac{13(1+i)}{2} \\ (b)\; \frac{13(1+i)}{4}\\ (c)\;\frac{13(1-i)}{2} \\ (d)\;\frac{15(1+i)}{2}\end{array}$

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