
How do you divide $\dfrac{8i}{-1+3i}$?
Answer
449.4k+ views
Hint: We are given a fraction, $\dfrac{8i}{-1+3i}$ and are asked to divide it. To solve this question, we will see what we have in the numerator and the denominator. We will learn about complex numbers, we learn what it stands for here. We will learn about conjugates of any term, then we will understand what the ways to divide 2 complex numbers are.
Complete step-by-step answer:
We are given a fraction $\dfrac{8i}{-1+3i}$ and are asked to divide them. Here, the numerator is 8i and the denominator is -1+3i. We have been given complex numbers.
So, we should understand that complex numbers are the numbers of the form a+ ib, where i stands for $\sqrt{-1}$ and is called iota. The conjugate of any complex number is the mirror image of it along the x-axis.
To divide the complex numbers, we will follow the following steps.
Step 1: We will find the conjugate of the denominator.
Step 2: We will multiply the numerator and the denominator of the given fraction by the conjugate of the denominator.
Step 3: We will then distribute the terms, we will find the product for the terms in the numerator as well as the denominator.
Step 4: We will simplify the power of i, always remember that ${{i}^{2}}=-1$.
Step 5: We will then group the like terms, that is, we will group the terms with iota together and the group the terms with constants together.
Step 6: We will simplify our answer in the standard complex form of a+ ib.
Now, we have $\dfrac{8i}{-1+3i}$. Here, the denominator is -1+3i. Now, we will find the conjugate of -1+ 3i, which is -1- 3i.
Now, we will multiply the numerator and the denominator with -1- 3i. So, we will get,
$\Rightarrow \dfrac{8i}{-1+3i}=\dfrac{8i}{-1+3i}\times \dfrac{-1-3i}{-1-3i}$
On simplifying, we get,
$\Rightarrow \dfrac{-8i+24}{1+3i-3i-9{{i}^{2}}}$
Now, we can cancel the terms. We will also use the property ${{i}^{2}}=-1$.
So, we get,
$\Rightarrow \dfrac{-8i+24}{1+9}$
Simplifying, we get,
$\Rightarrow \dfrac{24-8i}{10}$
Now, reducing it to the standard form, we get,
$\Rightarrow \dfrac{24}{10}-\dfrac{8}{10}i$
On simplifying further, we get,
$\Rightarrow \dfrac{12}{5}-\dfrac{4}{5}i$
Note: When we have complex numbers, we cannot simply add the constant terms and the terms with iota together, just like we cannot add constants and variables together. For example adding x+2 as 2x is wrong. Similarly adding 2i+ 2 as 4i is also wrong. One should remember to always add the like terms together. Also, we need to keep in the value of ${{i}^{2}}$ which is equal to -1 in order to simplify our solution.
Complete step-by-step answer:
We are given a fraction $\dfrac{8i}{-1+3i}$ and are asked to divide them. Here, the numerator is 8i and the denominator is -1+3i. We have been given complex numbers.
So, we should understand that complex numbers are the numbers of the form a+ ib, where i stands for $\sqrt{-1}$ and is called iota. The conjugate of any complex number is the mirror image of it along the x-axis.
To divide the complex numbers, we will follow the following steps.
Step 1: We will find the conjugate of the denominator.
Step 2: We will multiply the numerator and the denominator of the given fraction by the conjugate of the denominator.
Step 3: We will then distribute the terms, we will find the product for the terms in the numerator as well as the denominator.
Step 4: We will simplify the power of i, always remember that ${{i}^{2}}=-1$.
Step 5: We will then group the like terms, that is, we will group the terms with iota together and the group the terms with constants together.
Step 6: We will simplify our answer in the standard complex form of a+ ib.
Now, we have $\dfrac{8i}{-1+3i}$. Here, the denominator is -1+3i. Now, we will find the conjugate of -1+ 3i, which is -1- 3i.
Now, we will multiply the numerator and the denominator with -1- 3i. So, we will get,
$\Rightarrow \dfrac{8i}{-1+3i}=\dfrac{8i}{-1+3i}\times \dfrac{-1-3i}{-1-3i}$
On simplifying, we get,
$\Rightarrow \dfrac{-8i+24}{1+3i-3i-9{{i}^{2}}}$
Now, we can cancel the terms. We will also use the property ${{i}^{2}}=-1$.
So, we get,
$\Rightarrow \dfrac{-8i+24}{1+9}$
Simplifying, we get,
$\Rightarrow \dfrac{24-8i}{10}$
Now, reducing it to the standard form, we get,
$\Rightarrow \dfrac{24}{10}-\dfrac{8}{10}i$
On simplifying further, we get,
$\Rightarrow \dfrac{12}{5}-\dfrac{4}{5}i$
Note: When we have complex numbers, we cannot simply add the constant terms and the terms with iota together, just like we cannot add constants and variables together. For example adding x+2 as 2x is wrong. Similarly adding 2i+ 2 as 4i is also wrong. One should remember to always add the like terms together. Also, we need to keep in the value of ${{i}^{2}}$ which is equal to -1 in order to simplify our solution.
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