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Question: How do you simplify \(\dfrac{7+4i}{2-3i}\) ?...

How do you simplify 7+4i23i\dfrac{7+4i}{2-3i} ?

Explanation

Solution

As we know complex numbers have it’s real as well as an imaginary part and it is in the form a+bia+bi where ‘a’ is the real part and ‘b’ is the imaginary part. So, in this question, for division of complex numbers we will multiply the whole problem with the complex conjugate of 23i2-3i so that the denominator is real. Then we will distribute in numerator and denominator in order to remove the parentheses. After this, further simplification will be done by substituting the value of (i2)\left( {{i}^{2}} \right) where the value of i2{{i}^{2}} is -1.

Complete step by step answer:
Now, let’s solve this complex problem.
First, write the problem given in question:
7+4i23i\Rightarrow \dfrac{7+4i}{2-3i}
Now, the very first step is to multiply the whole problem with the complex conjugate. In order 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.
7+4i23i×(2+3i)(2+3i)\Rightarrow \dfrac{7+4i}{2-3i}\times \dfrac{\left( 2+3i \right)}{\left( 2+3i \right)}
Second step is to distribute in both numerator and denominator to remove the parentheses. In the denominator we can apply the identity (a+b)(ab)=a2b2\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}.
14+21i+8i+12i222(3i)2\Rightarrow \dfrac{14+21i+8i+12{{i}^{2}}}{{{2}^{2}}-{{\left( 3i \right)}^{2}}}
Now, substitute the value of i2{{i}^{2}} as -1.
We will get:
14+21i+8i124+9\Rightarrow \dfrac{14+21i+8i-12}{4+9}
On further calculation:
14+29i1213\Rightarrow \dfrac{14+29i-12}{13}
2+29i13\Rightarrow \dfrac{2+29i}{13}
Now, we need to write our answer in the form of a+bia+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+bia+bi form:
213+29i13\Rightarrow \dfrac{2}{13}+\dfrac{29i}{13}
Where real part is: 213\dfrac{2}{13} and imaginary part is 2913\dfrac{29}{13}
If further it can be reduced, then reduce it otherwise leave the answer as it is.

Note: Students usually put the value of ii as 1\sqrt{-1}. Remember that it will create a major mistake in your solution. So don’t make such silly mistakes ever. 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.