
How do you divide $\dfrac{4-3i}{5+5i}$?
Answer
556.2k+ views
Hint: To divide the given fraction which is in the form of a complex number, we have to convert the denominator of the fraction into a real number by multiplying the number by the complex conjugate of the denominator.
Complete answer:
We have the given number as:
$\Rightarrow \dfrac{4-3i}{5+5i}$
Now to get the denominator into the real form we must multiply it with the complex conjugate,
We know that for a complex number $a+bi$, the complex conjugate for that number is $a-bi$, where the terms $a$and $b$are real numbers.
Therefore, the complex conjugate of the number $5+5i$ will be $5-5i$, therefore we will multiply the numerator and denominator of the number with the term $5-5i$.
On multiplying we get:
$\Rightarrow \dfrac{4-3i}{5+5i}\times \dfrac{(5-5i)}{(5-5i)}$
On multiplying we get:
$\Rightarrow \dfrac{(4-3i)(5-5i)}{(5+5i)(5-5i)}$
Now on distributing the terms, we get:
$\Rightarrow \dfrac{4\times 5-4\times 5i-3i\times 5+3i\times 5i}{5\times 5-5\times 5i+5i\times 5-5i\times 5i}$
On multiplying the terms, we get:
$\Rightarrow \dfrac{20-35i+15{{i}^{2}}}{25-25{{i}^{2}}}$
Now we know that the value of ${{i}^{2}}=-1$, therefore on substituting in the equation, we get-
$\Rightarrow \dfrac{20-35i+15(-1)}{25-25(-1)}$
On simplifying we get:
$\Rightarrow \dfrac{20-15-35i}{25+25}$
$\Rightarrow \dfrac{5-35i}{50}$
Now on splitting the fraction, we get:
$\Rightarrow \dfrac{5}{50}-\dfrac{35i}{50}$
On simplifying, we get:
$\Rightarrow \dfrac{1}{10}-\dfrac{7}{10}i$, which is the final answer in the form of $a+bi$ where $a=\dfrac{1}{10}$ and $b=-\dfrac{7}{10}$ which are both real numbers.
Note:
it is to be remembered that whenever a complex number is multiplied with its complex conjugate, the complex part of the number is eliminated. It can be proved as:
$(a+bi)(a-bi)={{a}^{2}}-abi+abi-{{b}^{2}}{{i}^{2}}$
On simplifying we get:
$(a+bi)(a-bi)={{a}^{2}}-{{b}^{2}}{{i}^{2}}$
Now since the ${{i}^{2}}=-1$ we can write the term as:
$(a+bi)(a-bi)={{a}^{2}}+{{b}^{2}}$.
When we multiply and divide a number by a same number, the value of the fraction does not change, this is the reason we multiplied and divided the original term by the complex conjugate.
Complete answer:
We have the given number as:
$\Rightarrow \dfrac{4-3i}{5+5i}$
Now to get the denominator into the real form we must multiply it with the complex conjugate,
We know that for a complex number $a+bi$, the complex conjugate for that number is $a-bi$, where the terms $a$and $b$are real numbers.
Therefore, the complex conjugate of the number $5+5i$ will be $5-5i$, therefore we will multiply the numerator and denominator of the number with the term $5-5i$.
On multiplying we get:
$\Rightarrow \dfrac{4-3i}{5+5i}\times \dfrac{(5-5i)}{(5-5i)}$
On multiplying we get:
$\Rightarrow \dfrac{(4-3i)(5-5i)}{(5+5i)(5-5i)}$
Now on distributing the terms, we get:
$\Rightarrow \dfrac{4\times 5-4\times 5i-3i\times 5+3i\times 5i}{5\times 5-5\times 5i+5i\times 5-5i\times 5i}$
On multiplying the terms, we get:
$\Rightarrow \dfrac{20-35i+15{{i}^{2}}}{25-25{{i}^{2}}}$
Now we know that the value of ${{i}^{2}}=-1$, therefore on substituting in the equation, we get-
$\Rightarrow \dfrac{20-35i+15(-1)}{25-25(-1)}$
On simplifying we get:
$\Rightarrow \dfrac{20-15-35i}{25+25}$
$\Rightarrow \dfrac{5-35i}{50}$
Now on splitting the fraction, we get:
$\Rightarrow \dfrac{5}{50}-\dfrac{35i}{50}$
On simplifying, we get:
$\Rightarrow \dfrac{1}{10}-\dfrac{7}{10}i$, which is the final answer in the form of $a+bi$ where $a=\dfrac{1}{10}$ and $b=-\dfrac{7}{10}$ which are both real numbers.
Note:
it is to be remembered that whenever a complex number is multiplied with its complex conjugate, the complex part of the number is eliminated. It can be proved as:
$(a+bi)(a-bi)={{a}^{2}}-abi+abi-{{b}^{2}}{{i}^{2}}$
On simplifying we get:
$(a+bi)(a-bi)={{a}^{2}}-{{b}^{2}}{{i}^{2}}$
Now since the ${{i}^{2}}=-1$ we can write the term as:
$(a+bi)(a-bi)={{a}^{2}}+{{b}^{2}}$.
When we multiply and divide a number by a same number, the value of the fraction does not change, this is the reason we multiplied and divided the original term by the complex conjugate.
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