Question
Answer
$$\dfrac{(-1+i)^2}{-5+i}=\dfrac{-1+5i}{13}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{(-1+i)^2}{-5+i}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1-2i+i^2}{-5+i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{1-2i-1}{-5+i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}-\frac{2i}{-5+i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-1+5i}{13}\end{aligned} $$ | |
| ① | Find $ \left(-1+i\right)^2 $ in two steps. S1: Change all signs inside bracket. S2: Apply 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}{ i }$. $$ \begin{aligned}\left(-1+i\right)^2& \xlongequal{ S1 } \left(1-i\right)^2 \xlongequal{ S2 } \color{blue}{1^2} -2 \cdot 1 \cdot i + \color{red}{i^2} = \\[1 em] & = 1-2i+i^2\end{aligned} $$ |
| ② | $$ i^2 = -1 $$ |
| ③ | Simplify numerator $$ \, \color{blue}{ \cancel{1}} \,-2i \, \color{blue}{ -\cancel{1}} \, = -2i $$ |
| ④ | Divide $ \, -2i \, $ by $ \, -5+i \, $ to get $\,\, \dfrac{-1+5i}{13} $. ( view steps ) |
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