Answer
$$(6+2i)\cdot(6-2i)-(3-4i)\cdot(3+4i)=15$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(6+2i)\cdot(6-2i)-(3-4i)\cdot(3+4i)& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}36-12i+12i-4i^2-(9+12i-12i-16i^2) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}-4i^2+36-(-16i^2+9) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}4+36-(16+9) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}40-25 \xlongequal{ } \\[1 em] & \xlongequal{ }15\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{6+2i}\right) $ by each term in $ \left( 6-2i\right) $. $$ \left( \color{blue}{6+2i}\right) \cdot \left( 6-2i\right) = 36 -\cancel{12i}+ \cancel{12i}-4i^2 $$Multiply each term of $ \left( \color{blue}{3-4i}\right) $ by each term in $ \left( 3+4i\right) $. $$ \left( \color{blue}{3-4i}\right) \cdot \left( 3+4i\right) = 9+ \cancel{12i} -\cancel{12i}-16i^2 $$ |
| ② | Combine like terms: $$ 36 \, \color{blue}{ -\cancel{12i}} \,+ \, \color{blue}{ \cancel{12i}} \,-4i^2 = -4i^2+36 $$Combine like terms: $$ 9+ \, \color{blue}{ \cancel{12i}} \, \, \color{blue}{ -\cancel{12i}} \,-16i^2 = -16i^2+9 $$ |
| ③ | $$ -4i^2 = -4 \cdot (-1) = 4 $$$$ -16i^2 = -16 \cdot (-1) = 16 $$ |
| ④ | Combine like terms: $$ \color{blue}{4} + \color{blue}{36} = \color{blue}{40} $$Combine like terms: $$ \color{blue}{16} + \color{blue}{9} = \color{blue}{25} $$ |
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