Solveeit Logo
Login

Question

Question: How do you simplify \(\dfrac{{3 - 2i}}{{3 + 2i}}\) ?...

How do you simplify 3−2i3+2i\dfrac{{3 - 2i}}{{3 + 2i}} ?

Explanation

Solution

For solving this particular question, find the conjugate of the complex denominator. After multiplying the conjugate with the denominator and numerator , we get the real number in the denominator then use identities and simplify the given expression. At last, split the solution into its real and imaginary part.

Formula Used:
For solving the particular problem, we used two identities that are as follows.
(a−b)(a+b)=a2−b2(a - b)(a + b) = {a^2} - {b^2}
And
(a−b)2=a2−2ab+b2{(a - b)^2} = {a^2} - 2ab + {b^2}
We also used i2=−1{i^2} = - 1 .

Complete step by step answer:
We have to simplify the following expression,
3−2i3+2i\dfrac{{3 - 2i}}{{3 + 2i}} ,
Now, we have to find the conjugate of the denominator .So, we can simplify it . Since the denominator that is 3+2i3 + 2i is a complex number therefore its conjugate is 3−2i3 - 2i . Now multiply the conjugate with the denominator as well as with the numerator , by multiplying the complex number with its conjugate gives a real number . Now, eliminate the complex number from the denominator of the given expression,
⇒3−2i3+2i×3−2i3−2i\Rightarrow \dfrac{{3 - 2i}}{{3 + 2i}} \times \dfrac{{3 - 2i}}{{3 - 2i}}

Now using the identity (a−b)(a+b)=a2−b2(a - b)(a + b) = {a^2} - {b^2} ,
We can write ,
⇒(3−2i)232−(2i)2\Rightarrow \dfrac{{{{(3 - 2i)}^2}}}{{{3^2} - {{(2i)}^2}}}
Now using the identity (a−b)2=a2−2ab+b2{(a - b)^2} = {a^2} - 2ab + {b^2} ,
We can write ,
⇒9+4(i)2−2(3)(2i)9−4(i)2\Rightarrow \dfrac{{9 + 4{{(i)}^2} - 2(3)(2i)}}{{9 - 4{{(i)}^2}}}
Since we know that i2=−1{i^2} = - 1 ,
⇒9+4(−1)−2(3)(2i)9−4(−1) ⇒5−12i13 \Rightarrow \dfrac{{9 + 4( - 1) - 2(3)(2i)}}{{9 - 4( - 1)}} \\\ \Rightarrow \dfrac{{5 - 12i}}{{13}} \\\

∴513−1213i\therefore\dfrac{5}{{13}} - \dfrac{{12}}{{13}}i is the required solution.

Note: When we multiply complex numbers, multiplication is similar to the multiplication of two real numbers. In fact, multiplying two complex numbers is way similar to multiplying two binomials . Every term present in the first number must be multiplied with every term present in the other number. Any number which can be represented as a+bia + bi where aa and bb are real numbers then it is complex. And it exists on the complex plane.