
How do you divide $\dfrac{4+5i}{2-3i}$ ?
Answer
564.6k+ views
Hint: A complex number has it’s real as well as an imaginary part and it is in the form $a+bi$ where ‘a’ is the real part and ‘b’ is the imaginary part. In this question, for division of complex numbers we will multiply the whole problem with the complex conjugate of $2-3i$ so that the denominator is real. Then we distribute in numerator and denominator in order to remove the parentheses. Then further simplification will be done by substituting the value of $\left( {{i}^{2}} \right)$ where the value of ${{i}^{2}}$ is -1.
Complete step by step answer:
Now, let’s solve this complex problem.
First, write the problem given in question:
$\Rightarrow \dfrac{4+5i}{2-3i}$
Now, the first step is to multiply the whole problem with the complex conjugate. To find the conjugate of a complex number you have to change the sign between the two terms in the denominator and multiply with both numerator and denominator.
Let’s see how this needs to be done.
$\Rightarrow \dfrac{4+5i}{2-3i}\times \dfrac{\left( 2+3i \right)}{\left( 2+3i \right)}$
Second step is to distribute in both numerator and denominator in order to remove the parentheses. In the denominator we can apply the identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$.
$\Rightarrow \dfrac{8+12i+10i+15{{i}^{2}}}{{{2}^{2}}-{{\left( 3i \right)}^{2}}}$
Now, substitute the value of ${{i}^{2}}$ as -1.
We will get:
$\Rightarrow \dfrac{8+12i+10i-15}{4+9}$
On further calculation:
$\Rightarrow \dfrac{8+22i-15}{13}$
$\Rightarrow \dfrac{-7+22i}{13}$
Now, we need to write our answer in the form of $a+bi$ where ‘a’ is the real part of the complex number and ‘b’ is the imaginary part.
So, let’s write the answer in $a+bi$ form:
$\Rightarrow \dfrac{-7}{13}+\dfrac{22i}{13}$
Where real part is: $\dfrac{-7}{13}$ and imaginary part is $\dfrac{22}{13}$
If further it can be reduced, then reduce it otherwise leave the answer as it is.
Note: If you put the value of $i$ as $\sqrt{-1}$ then it will create a major mistake in your solution. So don’t make such silly mistakes. Complex conjugate is an important step to be solved. At the end, don’t forget to mention the real and imaginary parts of the complex number obtained because it carries some weightage.
Complete step by step answer:
Now, let’s solve this complex problem.
First, write the problem given in question:
$\Rightarrow \dfrac{4+5i}{2-3i}$
Now, the first step is to multiply the whole problem with the complex conjugate. To find the conjugate of a complex number you have to change the sign between the two terms in the denominator and multiply with both numerator and denominator.
Let’s see how this needs to be done.
$\Rightarrow \dfrac{4+5i}{2-3i}\times \dfrac{\left( 2+3i \right)}{\left( 2+3i \right)}$
Second step is to distribute in both numerator and denominator in order to remove the parentheses. In the denominator we can apply the identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$.
$\Rightarrow \dfrac{8+12i+10i+15{{i}^{2}}}{{{2}^{2}}-{{\left( 3i \right)}^{2}}}$
Now, substitute the value of ${{i}^{2}}$ as -1.
We will get:
$\Rightarrow \dfrac{8+12i+10i-15}{4+9}$
On further calculation:
$\Rightarrow \dfrac{8+22i-15}{13}$
$\Rightarrow \dfrac{-7+22i}{13}$
Now, we need to write our answer in the form of $a+bi$ where ‘a’ is the real part of the complex number and ‘b’ is the imaginary part.
So, let’s write the answer in $a+bi$ form:
$\Rightarrow \dfrac{-7}{13}+\dfrac{22i}{13}$
Where real part is: $\dfrac{-7}{13}$ and imaginary part is $\dfrac{22}{13}$
If further it can be reduced, then reduce it otherwise leave the answer as it is.
Note: If you put the value of $i$ as $\sqrt{-1}$ then it will create a major mistake in your solution. So don’t make such silly mistakes. Complex conjugate is an important step to be solved. At the end, don’t forget to mention the real and imaginary parts of the complex number obtained because it carries some weightage.
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