Question

How do I divide complex numbers in standard form?

Answer 1

Hint: We are asked to show how we divide the two complex numbers, to do so we will learn what are conjugate, we will learn how to multiply complex numbers, we will learn how iota (i) power behaves. We use ${{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1$ , then we will follow step to perform division in the complex, we will work with an example to check the better understanding of the method. We will learn.

Complete step by step answer:
Complex number are usually written as $a+ib$ , where ‘a’, ‘b’ are real and iota (i) is used to denote $\sqrt{-1}$ ,If we have term as $\dfrac{a+ib}{c+id}$ , it mean $a+ib$ is divided by $c+id$ to divide the complex number. We will follow the following step –
Step 1: Firstly we find the conjugate of the denominator .
Step 2: We will multiply numerator and denominator of the given fraction by the conjugate of the denominator.
Step 3: We will then distribute the term mean
We will produce the term in numerator as well as in denominator.
Step 4: We will simplify the power of I, always remember that ${{i}^{2}}$ is given as -1.
Step 5: We will then combine the like term that means we will iota involving terms with each other and only constant with each other.
Step 6: We will simply answer our answer lastly in standard complex form $a+ib$ .
We will learn better by one example.
Say we have $\dfrac{3+2i}{4-2i}$
We have a numerator at $3+2i$ and denominator as $4-2i$ .
So, step 1: We will find conjugate of $4-2i$ conjugate at $a+ib$ is simple, it is $a-ib$ hence conjugate of $4-2i=4-\left( -2i \right)$
$=4+2i$
Step 2: We multiply numerator and denominator by $4+2i$ .
So, $\dfrac{\left( 3+2i \right)}{\left( 4-2i \right)}\times \dfrac{\left( 4+2i \right)}{4+2i}$
Step 3: We will multiply terms in the numerator as well as in the denominator.
S, we get –
$\left( \dfrac{3+2i}{4-2i} \right)\times \left( \dfrac{4+2i}{4+2i} \right)=\dfrac{12+4{{i}^{2}}+6i+8i}{16-4{{i}^{2}}+8i-8i}$
Step 4: We simplify by using ${{i}^{2}}=-1$ so, we get –
$=\dfrac{12-4+6i+8i}{16-\left( -4 \right)+8i-8i}$
Now we combine like term, and simplify –
$=\dfrac{8+14i}{20}$
As $12-4=8,6i+8i=14i$ and $-8i+8i=0$
Now, we will reduce above term into the standard form $a+ib$
$=\dfrac{8+14i}{20}=\dfrac{8}{20}+\dfrac{14}{20}i$
We will reduce the term
$=\dfrac{2}{5}+\dfrac{7}{10}i$
Hence, we get $\dfrac{3+2i}{4-2i}=\dfrac{2}{5}+\dfrac{7}{10}i$
This is how we divide the term in the complex numbers.

Note: When we have complex term remember we cannot add just constant with the iota term, just like we cannot add variable with constant i.e $x+2=2x$ is wrong similarly $2i+2=4i$ is wrong, we can always add like term, we done addition of the simplification of iota. We need to be careful with calculation.