Question
Answer
$$(1+i)\cdot(1-i)\cdot\dfrac{1}{2}=1$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(1+i)\cdot(1-i)\cdot\frac{1}{2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(1-i+i-i^2)\cdot\frac{1}{2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(-i^2+1)\cdot\frac{1}{2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}(1+1)\cdot\frac{1}{2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}2\cdot\frac{1}{2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}1\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{1+i}\right) $ by each term in $ \left( 1-i\right) $. $$ \left( \color{blue}{1+i}\right) \cdot \left( 1-i\right) = 1 -\cancel{i}+ \cancel{i}-i^2 $$ |
| ② | Combine like terms: $$ 1 \, \color{blue}{ -\cancel{i}} \,+ \, \color{blue}{ \cancel{i}} \,-i^2 = -i^2+1 $$ |
| ③ | $$ -i^2 = -(-1) = 1 $$ |
| ④ | Combine like terms: $$ \color{blue}{1} + \color{blue}{1} = \color{blue}{2} $$ |
| ⑤ | Multiply $2$ by $ \dfrac{1}{2} $ to get $ 1$. Step 1: Write $ 2 $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Cancel $ \color{blue}{ 2 } $ in first and second fraction. Step 3: Multiply numerators and denominators. $$ \begin{aligned} 2 \cdot \frac{1}{2} & \xlongequal{\text{Step 1}} \frac{2}{\color{red}{1}} \cdot \frac{1}{2} \xlongequal{\text{Step 2}} \frac{\color{blue}{1}}{1} \cdot \frac{1}{\color{blue}{1}} = \\[1ex] &= \frac{1}{1} =1 \end{aligned} $$ |
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