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Question

Question: How do you simplify the complex number \(\dfrac{5}{{2 + 3i}}\)?...

How do you simplify the complex number 52+3i\dfrac{5}{{2 + 3i}}?

Explanation

Solution

In this question, we need to simplify the given complex number 52+3i\dfrac{5}{{2 + 3i}}. To simplify this, we are going to use the method of rationalization. So, we will multiply and divide by 23i2 - 3i and then simplify the numerator and denominator of the above expression. After that we will obtain a simplified form. The power of ii(iota) that we require in solving the above problem is i2=1{i^2} = - 1.

Complete step-by-step solution:
Given the complex number of the form 52+3i\dfrac{5}{{2 + 3i}} …… (1)
We are asked to simplify the above complex number given in the equation (1).
Now to simplify the above complex number, we will use the method of rationalization.
So we are going to rationalize the above given complex number.
So to do this, we are going to multiply and divide the equation (1) by 23i2 - 3i, we get,
52+3i×23i23i\Rightarrow \dfrac{5}{{2 + 3i}} \times \dfrac{{2 - 3i}}{{2 - 3i}}
In the denominator we are multiplying 2+3i2 + 3i with 23i2 - 3i and in the numerator we are multiplying 5 with 23i2 - 3i.
Hence we get,
5(23i)(2+3i)(23i)\Rightarrow \dfrac{{5(2 - 3i)}}{{(2 + 3i)(2 - 3i)}} …… (2)
Note that in the denominator above we have a complex number of the form (a+b)(ab)(a + b)(a - b).
So we are going to use the identity, (a+b)(ab)=a2b2(a + b)(a - b) = {a^2} - {b^2}
Here we have the denominator as (2+3i)(23i)(2 + 3i)(2 - 3i)
Note that a=2a = 2 and b=3ib = 3i
Now using the above identity we get,
(2+3i)(23i)=22(3i)2(2 + 3i)(2 - 3i) = {2^2} - {(3i)^2}
(2+3i)(23i)=49i2\Rightarrow (2 + 3i)(2 - 3i) = 4 - 9{i^2}
We know that i=1i = \sqrt { - 1} . Squaring on both sides we get, i2=1{i^2} = - 1.
Substituting this we get,
(2+3i)(23i)=49(1)\Rightarrow (2 + 3i)(2 - 3i) = 4 - 9( - 1)
(2+3i)(23i)=4+9\Rightarrow (2 + 3i)(2 - 3i) = 4 + 9
(2+3i)(23i)=13\Rightarrow (2 + 3i)(2 - 3i) = 13
Hence the equation (2) becomes,
5(23i)13\Rightarrow \dfrac{{5(2 - 3i)}}{{13}}
Now multiplying the terms in the numerator we get,
5×25×3i13\Rightarrow \dfrac{{5 \times 2 - 5 \times 3i}}{{13}}
1015i13\Rightarrow \dfrac{{10 - 15i}}{{13}}
This can also be written as,
10131513i\Rightarrow \dfrac{{10}}{{13}} - \dfrac{{15}}{{13}}i
Hence the simplified form of the given complex number is 10131513i\dfrac{{10}}{{13}} - \dfrac{{15}}{{13}}i.

Note: Students must know how to rationalize the complex number and simplify such problems. The principle of rationalization is that we are going to multiply and divide the conjugate of the denominator and then simplify. Note that the conjugate of a+iba + ib is aiba – ib . Also whenever we see any complex number in the denominator either with positive or negative sign, then rationalization is the method we need to use and simplify the complex number.
Students may make mistakes while writing the value of i2{i^2} as 1 instead of -1. So make sure to remember this value to solve complex numbers.