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Question

Question: How do you divide \(\dfrac{4-3i}{5+5i}\)?...

How do you divide 43i5+5i\dfrac{4-3i}{5+5i}?

Explanation

Solution

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:
43i5+5i\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+bia+bi, the complex conjugate for that number is abia-bi, where the terms aaand bbare real numbers.
Therefore, the complex conjugate of the number 5+5i5+5i will be 55i5-5i, therefore we will multiply the numerator and denominator of the number with the term 55i5-5i.
On multiplying we get:
43i5+5i×(55i)(55i)\Rightarrow \dfrac{4-3i}{5+5i}\times \dfrac{(5-5i)}{(5-5i)}
On multiplying we get:
(43i)(55i)(5+5i)(55i)\Rightarrow \dfrac{(4-3i)(5-5i)}{(5+5i)(5-5i)}
Now on distributing the terms, we get:
4×54×5i3i×5+3i×5i5×55×5i+5i×55i×5i\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:
2035i+15i22525i2\Rightarrow \dfrac{20-35i+15{{i}^{2}}}{25-25{{i}^{2}}}
Now we know that the value of i2=1{{i}^{2}}=-1, therefore on substituting in the equation, we get-
2035i+15(1)2525(1)\Rightarrow \dfrac{20-35i+15(-1)}{25-25(-1)}
On simplifying we get:
201535i25+25\Rightarrow \dfrac{20-15-35i}{25+25}
535i50\Rightarrow \dfrac{5-35i}{50}
Now on splitting the fraction, we get:
55035i50\Rightarrow \dfrac{5}{50}-\dfrac{35i}{50}
On simplifying, we get:
110710i\Rightarrow \dfrac{1}{10}-\dfrac{7}{10}i, which is the final answer in the form of a+bia+bi where a=110a=\dfrac{1}{10} and b=710b=-\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)(abi)=a2abi+abib2i2(a+bi)(a-bi)={{a}^{2}}-abi+abi-{{b}^{2}}{{i}^{2}}
On simplifying we get:
(a+bi)(abi)=a2b2i2(a+bi)(a-bi)={{a}^{2}}-{{b}^{2}}{{i}^{2}}
Now since the i2=1{{i}^{2}}=-1 we can write the term as:
(a+bi)(abi)=a2+b2(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.