Answer
396.9k+ views
Hint:
Any number in the form of \[a+bi\] is known as a complex number. Its conjugate complex number is \[a-bi\]. Product of a complex and its conjugate complex number is the sum of squares of a and b, that is \[{{(a)}^{2}}+{{(b)}^{2}}\]. Most importantly, the square of \[i\] is equal to -1.
Complete step by step answer:
In the given question, the complex number is \[\dfrac{2-3i}{1-2i}\]. While simplifying such types of complex numbers, we will eliminate the complex number from the denominator by multiplying both the numerator and denominator with the conjugate complex number of the denominator.
Here the denominator complex number is \[(1-2i)\]. The conjugate complex number of the denominator \[(1-2i)\] is \[\left( 1+2i \right)\]. So, on multiplying both the numerator and denominator with \[\left( 1+2i \right)\] we get
\[\Rightarrow \dfrac{(2-3i)(1+2i)}{(1-2i)(1+2i)}\]
We know that \[\left( a+bi \right)(a-bi)={{(a)}^{2}}+{{(b)}^{2}}\]. On multiplying two constants we get a constant; on multiplying a constant (say c) with iota \[i\] gives \[ci\] and on multiplying two iota terms (say \[ci\], \[di\] ) we get \[{{i}^{2}}\] term with its coefficient \[c\times d\]. So, we can write
\[\Rightarrow \dfrac{(2+4i-3i-6{{i}^{2}})}{({{1}^{2}}-{{(2i)}^{2}})}\]
We know that square of \[i\] is equal to -1. On substituting -1 and subtraction of \[3i\] from \[4i\] we get
\[\begin{align}
& \Rightarrow \dfrac{(2+i-6(-1))}{(1-4(-1))} \\
& \Rightarrow \dfrac{(2+i+6)}{(1+4)} \\
\end{align}\]
Now on simple addition we get
\[\Rightarrow \dfrac{(8+i)}{5}\]
Hence the simplified form of \[\dfrac{2-3i}{1-2i}\] is \[\dfrac{8+i}{5}\].
Note:
If we want to simplify an expression, it is always important to keep in mind what we mean when we say ’simplify’. Typically, in the case of complex numbers, we aim to never have a complex number in the denominator of any term. To accomplish this, we multiply \[a+bi\] by its complex conjugate \[a-bi\] , where we end up with a real number \[{{(a)}^{2}}+{{(b)}^{2}}\]. This will allow us to simplify the complex nature out of a denominator.
Any number in the form of \[a+bi\] is known as a complex number. Its conjugate complex number is \[a-bi\]. Product of a complex and its conjugate complex number is the sum of squares of a and b, that is \[{{(a)}^{2}}+{{(b)}^{2}}\]. Most importantly, the square of \[i\] is equal to -1.
Complete step by step answer:
In the given question, the complex number is \[\dfrac{2-3i}{1-2i}\]. While simplifying such types of complex numbers, we will eliminate the complex number from the denominator by multiplying both the numerator and denominator with the conjugate complex number of the denominator.
Here the denominator complex number is \[(1-2i)\]. The conjugate complex number of the denominator \[(1-2i)\] is \[\left( 1+2i \right)\]. So, on multiplying both the numerator and denominator with \[\left( 1+2i \right)\] we get
\[\Rightarrow \dfrac{(2-3i)(1+2i)}{(1-2i)(1+2i)}\]
We know that \[\left( a+bi \right)(a-bi)={{(a)}^{2}}+{{(b)}^{2}}\]. On multiplying two constants we get a constant; on multiplying a constant (say c) with iota \[i\] gives \[ci\] and on multiplying two iota terms (say \[ci\], \[di\] ) we get \[{{i}^{2}}\] term with its coefficient \[c\times d\]. So, we can write
\[\Rightarrow \dfrac{(2+4i-3i-6{{i}^{2}})}{({{1}^{2}}-{{(2i)}^{2}})}\]
We know that square of \[i\] is equal to -1. On substituting -1 and subtraction of \[3i\] from \[4i\] we get
\[\begin{align}
& \Rightarrow \dfrac{(2+i-6(-1))}{(1-4(-1))} \\
& \Rightarrow \dfrac{(2+i+6)}{(1+4)} \\
\end{align}\]
Now on simple addition we get
\[\Rightarrow \dfrac{(8+i)}{5}\]
Hence the simplified form of \[\dfrac{2-3i}{1-2i}\] is \[\dfrac{8+i}{5}\].
Note:
If we want to simplify an expression, it is always important to keep in mind what we mean when we say ’simplify’. Typically, in the case of complex numbers, we aim to never have a complex number in the denominator of any term. To accomplish this, we multiply \[a+bi\] by its complex conjugate \[a-bi\] , where we end up with a real number \[{{(a)}^{2}}+{{(b)}^{2}}\]. This will allow us to simplify the complex nature out of a denominator.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)