Question

How do you divide \[\dfrac{2+5i}{5+2i}?\]

Answer 1

Hint: We are given \[\dfrac{2+5i}{5+2i}\] and we are asked to divide the following term. To divide we will learn about the complex number and how they are usually written and how multiplication and division are done in these. We will rationalize the denominator and then we use (a + b)(a – b) formula to simplify. We also need (x + y)(a + b) which is the basic knowledge of multiplication.

Complete step by step answer:
We are given the term as \[\dfrac{2+5i}{5+2i}\] and we are asked to solve this. Here, the division of 2 + 5i with 5 + 2i is the same as the simplification of \[\dfrac{2+5i}{5+2i}.\] So, we basically simplify our given expression. The complex number is basically written as a + ib where a and b are real numbers and i (iota) is used to denote \[\sqrt{-1}\] i.e. \[\sqrt{-1}=i.\] Now to simplify our rational form, we basically have to change the fraction into the simple fraction a + ib form. So, we have to remove the complex term from the denominator. Now, as we have 5 + 2i in the denominator, so we multiply and divide by its conjugate 5 – 2i, so we will get
\[\dfrac{2+5i}{5+2i}=\dfrac{2+5i}{5+2i}\times \dfrac{5-2i}{5-2i}\]
Now we will simplify the numerator and denominator one by one. On simplifying the numerator first, we get,
\[\left( 2+5i \right)\left( 5-2i \right)=2\left( 5-2i \right)+5i\left( 5-2i \right)\]
\[\Rightarrow \left( 2+5i \right)\left( 5-2i \right)=2\times 5-2\times 2i+5i\times 5+5i\times \left( -2i \right)\]
\[\Rightarrow \left( 2+5i \right)\left( 5-2i \right)=10-4i+25i-10\left( {{i}^{2}} \right)\]
As \[{{i}^{2}}=-1,\] so we get,
\[\Rightarrow \left( 2+5i \right)\left( 5-2i \right)=10+21i+10\]
\[\Rightarrow \left( 2+5i \right)\left( 5-2i \right)=20+21i\]
So, we get,
\[\left( 2+5i \right)\left( 5-2i \right)=20+21i\]
Now, simplifying the denominator, we get,
\[\left( 5+2i \right)\left( 5-2i \right)={{5}^{2}}-{{\left( 2i \right)}^{2}}\]
By using \[\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\] we get,
\[\Rightarrow \left( 5+2i \right)\left( 5-2i \right)=25-4{{i}^{2}}\]
As \[{{i}^{2}}=1,\] so we get,
\[\Rightarrow \left( 5+2i \right)\left( 5-2i \right)=25+4\]
\[\Rightarrow \left( 5+2i \right)\left( 5-2i \right)=29\]
So, we get,
\[\dfrac{2+5i}{5+2i}=\dfrac{20+21i}{29}\]
\[\Rightarrow \dfrac{2+5i}{5+2i}=\dfrac{20}{29}+\dfrac{21i}{29}\]
So, our answer is \[\dfrac{20}{29}+\dfrac{21i}{29}.\]

Note:
 While simplifying, always remember that in the product, we multiply each term of the first bracket by each term of the other bracket. Do not make errors like (a + b)(c + d) = ac + b which is wrong. Also, while simplifying things in addition or subtraction, we have to apply the sign that is outside the bracket to all terms of the bracket. Do not make mistakes like (a + b) – (c + d) = a + b – c + d which is wrong. It should be like (a + b) – (c + d) = a + b – c – d.