Question
Answer
$$(1-4i)\dfrac{1+4i}{(1+4i)^2}=\dfrac{-8i-15}{17}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(1-4i)\frac{1+4i}{(1+4i)^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(1-4i)\frac{1+4i}{1+8i+16i^2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(1-4i)\frac{1+4i}{1+8i-16} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}(1-4i)\frac{1+4i}{8i-15} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}(1-4i)\frac{1-4i}{17} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{16i^2-8i+1}{17} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}\frac{-16-8i+1}{17} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}\frac{-8i-15}{17}\end{aligned} $$ | |
| ① | Find $ \left(1+4i\right)^2 $ using formula. $$ (A + B)^2 = \color{blue}{A^2} + 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ 1 } $ and $ B = \color{red}{ 4i }$. $$ \begin{aligned}\left(1+4i\right)^2 = \color{blue}{1^2} +2 \cdot 1 \cdot 4i + \color{red}{\left( 4i \right)^2} = 1+8i+16i^2\end{aligned} $$ |
| ② | $$ 16i^2 = 16 \cdot (-1) = -16 $$ |
| ③ | Combine like terms: $$ \color{blue}{1} +8i \color{blue}{-16} = 8i \color{blue}{-15} $$ |
| ④ | Divide $ \, 1+4i \, $ by $ \, -15+8i \, $ to get $\,\, \dfrac{1-4i}{17} $. ( view steps ) |
| ⑤ | Multiply $1-4i$ by $ \dfrac{1-4i}{17} $ to get $ \dfrac{16i^2-8i+1}{17} $. Step 1: Write $ 1-4i $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} 1-4i \cdot \frac{1-4i}{17} & \xlongequal{\text{Step 1}} \frac{1-4i}{\color{red}{1}} \cdot \frac{1-4i}{17} \xlongequal{\text{Step 2}} \frac{ \left( 1-4i \right) \cdot \left( 1-4i \right) }{ 1 \cdot 17 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 1-4i-4i+16i^2 }{ 17 } = \frac{16i^2-8i+1}{17} \end{aligned} $$ |
| ⑥ | $$ 16i^2 = 16 \cdot (-1) = -16 $$ |
| ⑦ | Simplify numerator $$ \color{blue}{-16} -8i+ \color{blue}{1} = -8i \color{blue}{-15} $$ |
This page was created using
Complex Expression calculator