Solution 3.1:1d

We expand the expression by multiplying each term in the first bracket with every term in the second bracket:

\displaystyle \begin{align}(3-2i)(7+5i)&=3\cdot 7 + 3 \cdot 5i + \cdots\\ &=3\cdot 7 + 3 \cdot 5i-2i\cdot 7 -2i \cdot 5i.\end{align}

Then, we use that \displaystyle i^2=-1 and write the real and imaginary parts together:

\displaystyle \begin{align}(3-2i)(7+5i)&=21+15i-14i-10i^2\\ &=21+15i-14i-10\cdot(-1)\\ &=(21+10)+(15i-14i)\\ &=31+1i\\ &=31+i\end{align}