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Question: How do you write the following quotient in standard form\[\dfrac{8-7i}{1-2i}\]?...

How do you write the following quotient in standard form8−7i1−2i\dfrac{8-7i}{1-2i}?

Explanation

Solution

This question belongs to the topic of calculus. In this question, we are going to use the theories of complex numbers and are going to understand the term about iota. So that we can solve this question easily. We will first find the complex conjugate of the denominator of the given term 8−7i1−2i\dfrac{8-7i}{1-2i}. After that, we will multiply the term 8−7i1−2i\dfrac{8-7i}{1-2i} with the complex conjugate of the denominator. After simplifying the term, we will get the solution.

Complete step by step answer:
Let us solve this question.
In this question, it is asked to simplify the term 8−7i1−2i\dfrac{8-7i}{1-2i}. So, we will solve this term by multiplying the complex conjugate of denominator to numerator and denominator.
Let us first know about the complex conjugate.
Let us take a complex number.
If the complex number is x+iyx+iy, then the complex conjugate of the term x+iyx+iy will be x−iyx-iy.
So, we can say that the complex conjugate of the term 1−2i1-2i will be 1+2i1+2i.
Now, we will multiply the term 1+2i1+2i with the numerator and denominator of the term 8−7i1−2i\dfrac{8-7i}{1-2i}.
So, we can write
8−7i1−2i=8−7i1−2i×1+2i1+2i\dfrac{8-7i}{1-2i}=\dfrac{8-7i}{1-2i}\times \dfrac{1+2i}{1+2i}
The above equation can also be written as
⇒8−7i1−2i=(8−7i)(1+2i)(1−2i)(1+2i)\Rightarrow \dfrac{8-7i}{1-2i}=\dfrac{\left( 8-7i \right)\left( 1+2i \right)}{\left( 1-2i \right)\left( 1+2i \right)}
We are going to use the foil method here.
The foil method says that: (a+b)(c+d)=ac+ad+bc+bd\left( a+b \right)\left( c+d \right)=ac+ad+bc+bd
From foil method, we can say that
(a−b)(a+b)=a2−b2\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}
Using the above two formulas, we can say that
⇒8−7i1−2i=8×1+8×2i+(−7i)×1+(−7i)×2i12−(−2i)2\Rightarrow \dfrac{8-7i}{1-2i}=\dfrac{8\times 1+8\times 2i+\left( -7i \right)\times 1+\left( -7i \right)\times 2i}{{{1}^{2}}-{{\left( -2i \right)}^{2}}}
The above equation can also be written as
⇒8−7i1−2i=8+16i−7i−14i212−4i2\Rightarrow \dfrac{8-7i}{1-2i}=\dfrac{8+16i-7i-14{{i}^{2}}}{{{1}^{2}}-4{{i}^{2}}}
The above equation can also be written as
⇒8−7i1−2i=8+9i−14i21−4i2\Rightarrow \dfrac{8-7i}{1-2i}=\dfrac{8+9i-14{{i}^{2}}}{1-4{{i}^{2}}}
Now, we will get to know about iota.
The term iota can be written as ii.
The value of ii is −1\sqrt{-1}.
So, we can write
i1=i=−1{{i}^{1}}=i=\sqrt{-1} and i2=−1{{i}^{2}}=-1
Now, we will use them and solve the above equation. We can write the above equation as
⇒8−7i1−2i=8+9i−14(−1)1−4(−1)\Rightarrow \dfrac{8-7i}{1-2i}=\dfrac{8+9i-14\left( -1 \right)}{1-4\left( -1 \right)}
The above equation can also be written as
⇒8−7i1−2i=8+9i+141+4\Rightarrow \dfrac{8-7i}{1-2i}=\dfrac{8+9i+14}{1+4}
The above equation can also be written as
⇒8−7i1−2i=22+9i5\Rightarrow \dfrac{8-7i}{1-2i}=\dfrac{22+9i}{5}
Now, we will convert the term 22+9i5\dfrac{22+9i}{5} in the term of x+iyx+iy.

So, we can write the term8−7i1−2i\dfrac{8-7i}{1-2i} in standard form as 225+95i\dfrac{22}{5}+\dfrac{9}{5}i

Note: We should have a better knowledge of complex numbers to solve this type of question easily. Remember the formula of foil method.
The foil method says that:
If we multiply the terms (a+b) and (c+d), then the multiplied term will be ac+ad+bc+bd.
And, also remember that
(a−b)(a+b)=a2−b2\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}
And, don’t forget the values of iota and square of iota.
The value of iota is i=−1i=\sqrt{-1}
And the square of iota will be i2=−1{{i}^{2}}=-1.
These formulas and methods are very useful for this type of question. So, try to keep remembering these formulas and methods.