Question

How do you divide $ \dfrac{2+3i}{1+2i} $ ?

Answer 1

Hint: In this question, we need to divide two complex numbers of the form a+bi. For this, we need to remove the complex number from the denominator. We will multiply and divide the fraction by a certain complex number which will help in removing the complex number from the denominator. If a denominator has the form c+id, we multiply and divide the fraction by its conjugate which is c-id. Then we will apply the arithmetic property that $ \left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}} $ to simplify further. We will also use the property that $ {{i}^{2}}=-1 $ .

Complete step by step solution:
Here we have the fraction given as $ \dfrac{2+3i}{1+2i} $ .
We need to simplify it. As we cannot simply divide it, so we need to remove complex numbers from the denominator. We know that, when we have a denominator of the form c+id we need to multiply and divide the fraction by its conjugate.
Conjugate of c+id is given by c-id. Here we have the denominator as 1+2i. So the conjugate of 1+2i will be given by 1-2i. Therefore we need to multiply and divide the fraction by 1-2i we get \[\dfrac{2+3i}{1+2i}\times \dfrac{1-2i}{1-2i}\].
It can be written as \[\dfrac{\left( 2+3i \right)\left( 1-2i \right)}{\left( 1+2i \right)\left( 1-2i \right)}\].
Let us simplify it.
For numerator, let us open brackets and simplify the terms we get \[\dfrac{2\left( 1 \right)-2\left( 2i \right)+3i\left( 1 \right)-\left( 3i \right)\left( 2i \right)}{\left( 1+2i \right)\left( 1-2i \right)}\].
Simplifying the numerator we get \[\dfrac{2-4i+3i-6{{i}^{2}}}{\left( 1+2i \right)\left( 1-2i \right)}\].
We know that $ i=\sqrt{-1} $ . So squaring both sides we get $ {{i}^{2}}=-1 $ . So numerator becomes \[\dfrac{2-4i+3i-6\left( -1 \right)}{\left( 1+2i \right)\left( 1-2i \right)}\Rightarrow \dfrac{2-4i+3i-6}{\left( 1+2i \right)\left( 1-2i \right)}\].
Adding or subtracting the like terms we get \[\dfrac{8-i}{\left( 1+2i \right)\left( 1-2i \right)}\].
Now simplifying the denominator we can see that the denominator is of the form (a+b)(a-b). So we can use the property of arithmetic and write (a+b)(a-b) as $ {{a}^{2}}-{{b}^{2}} $ we get \[\dfrac{8-i}{{{\left( 1 \right)}^{2}}-{{\left( 2i \right)}^{2}}}\Rightarrow \dfrac{8-i}{1-4{{i}^{2}}}\].
Using $ {{i}^{2}}=-1 $ in the denominator we get \[\dfrac{8-i}{1-4\left( -1 \right)}\Rightarrow \dfrac{8-i}{1+4}\Rightarrow \dfrac{8-i}{5}\].
To get a simple complex number, separate terms we get \[\dfrac{8}{5}-\dfrac{i}{5}\].
Therefore $ \dfrac{2+3i}{1+2i} $ reduces to \[\dfrac{8}{5}-\dfrac{i}{5}\] which is the required answer.

Note:
 Students should carefully separate the terms while multiplying two complex numbers. Take care of the signs while solving the sum. Make sure to separate terms at the end to get a single complex number of the form a+ib.