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Division of complex numbers calculator
Use this online calculator to divide complex numbers.
The calculator shows a step-by-step, easy-to-understand solution on how the division was done.
Divide complex numbers:
$$ \dfrac{20+10i}{1-3i} $$solution
$$ \dfrac{20+10i}{1-3i} = -1+7i $$explanation
Step 1: Determine the conjugate of the denominator. ( to find the conjugate just change the sign of the imaginary part ).
In this example, the conjugate of $ \color{orangered}{ 1-3i }\, $ is $ \color{blue}{ 1+3i } $.
Step 2: Multiply both the numerator and denominator by the conjugate:
$$\begin{aligned} \frac{ 20+10i }{ 1-3i } &= \frac{ 20+10i }{ 1-3i } \cdot \frac{ \color{blue}{ 1+3i } }{ \color{blue}{ 1+3i } } = \\[1 em] &= \frac{ \left( 20+10i \right) \cdot \left( 1+3i \right) }{ \left( 1-3i \right) \cdot \left( 1+3i \right) } \end{aligned} $$Step 3: Simplify numerator and denominator (use $\color{blue}{i^2 = -1}$)
$$ \begin{aligned} \left( 20+10i \right) \cdot \left( 1+3i \right) &= 20 \cdot 1 + 20 \cdot \left(3 \,i \right) + \left( 10 \,i \right) \cdot \left(1 \right) + \left( 10 \,i \right) \cdot \left(3 \,i \right) = \\[1 em] &= 20 + 60 \, i + 10 \, i + 30 \color{blue}{(-1)} = \\[1 em] &= -10+70i\end{aligned} $$ $$ \begin{aligned} \left( 1-3i \right) \cdot \left( 1+3i \right) &= 1 \cdot 1 + 1 \cdot \left(3 \,i \right) + \left( -3 \,i \right) \cdot \left(1 \right) + \left( -3 \,i \right) \cdot \left(3 \,i \right) = \\[1 em] &= 1 + 3 \, i -3 \, i -9 \color{blue}{(-1)} = \\[1 em] &= 10\end{aligned} $$Step 4: Separate real and imaginary parts:
$$ \frac{ 20+10i }{ 1-3i } = \frac{ -10+70i }{ 10 } = \frac{ -10 }{ 10 } + \frac{ 70 }{ 10 } i= -1+7i $$How to divide complex numbers?
This calculator uses multiplication by conjugate to divide complex numbers.
Example 1:
$$ \frac{ 4 + 2i }{1 + i} $$We begin by multiplying numerator and denominator by complex conjugate of $ \color{purple}{1 + i} $.
$$ \frac{4 + 2i}{\color{purple}{1 + i}} \cdot \frac{\color{blue}{1 - i}}{\color{blue}{1 - i}} = \frac{(4+2i)(1-i)}{(1+i)(1-i)}$$Then we expand and simplify both products. Keep in mind that $ i^2 = -1 $.
$$ \begin{aligned} \frac{(4+2i)(1-i)}{(1+i)(1-i)} &= \frac{4 - 4i + 2i - 2\color{blue}{i^2}}{1+i-i-i^2} = \\[ 1 em] &= \frac{4 - 2i - 2\color{blue}{(-1)}}{1-\color{purple}{i^2}} = \\[ 1 em] &= \frac{4 - 2i + 2)}{1-\color{purple}{(-1)}} = \\[ 1 em] &= \frac{6 - 2i)}{2} \end{aligned} $$At the end we separate real and imaginary parts:
$$ \frac{6 - 2i}{2} = \frac{6}{2} - \frac{2}{2}i = 3 - i $$Example 2:
Divide $ 10 - 25i $ by $ 5i $
Although the complex conjugate of $ 5i $ is $-5i$, we can simplify division process by multiplying numerator and denominator with $ - i $.
$$ \begin{aligned} \frac{10-25i}{5i} &= \frac{10-25i}{5i} \cdot \frac{-i}{-i} = \\[1 em] &= \frac{(10-25i)(-i)}{(5i)(-i)}= \\[ 1 em] &= \frac{-10i + 25i^2}{-5i^2} = \\[ 1 em] &= \frac{-10i - 25}{5} = \\[ 1 em] &= \frac{-25}{5} + \frac{-10}{5} i= \\[ 1 em] &= -5 - 2 i= \\[ 1 em] \end{aligned} $$Example 3:
Divide $ 20 + 10i $ by $ 1 - 3i $
Please tell me how can I make this better.