Question

How do I find the quotient of two complex numbers in standard form?

Answer 1

Hint: We will first start by considering a few terms, then elaborate each and every term precisely. After that, mention exactly what we need to evaluate. Mention the generalised form too as well. Then apply the conditions and proceed.

Complete step by step solution:
Here we will start by considering two terms as $ {z_1} = {a_1} + {b_1}i $ and $ {z_2} = {a_2} + {b_2}i $ . Here we want to evaluate the value of the term \[q = \dfrac{{{z_1}}}{{{z_2}}} = \dfrac{{{a_1} + {b_1}i}}{{{a_1} + {b_1}i}}\] .
Generally, we write the same term in the form: $ q = A + Bi $ .
Now, here $ A,B $ are real numbers. For doing this, we must amplify the quotient by the conjugate of the denominator.
Hence, we can evaluate the equation as,
 \[
  q = \dfrac{{{z_1}}}{{{z_2}}}.\dfrac{{\overline {{z_2}} }}{{\overline {{z_2}} }} = \dfrac{{{a_1} + {b_1}i}}{{{a_2} + {b_2}i}}.\dfrac{{{a_2} + {b_2}i}}{{{a_2} + {b_2}i}} \\
  q = \dfrac{{({a_1}{a_2} + {b_1}{b_2}) + ({b_1}{a_2} - {b_2}{a_1})i}}{{{a^2}_2 + {b^2}_2}} \\
  q = \dfrac{{({a_1}{a_2} + {b_1}{b_2})}}{{{a^2}_2 + {b^2}_2}} + \dfrac{{({b_1}{a_2} - {b_2}{a_1})}}{{{a^2}_2 + {b^2}_2}}i \;
 \]
So, the correct answer is “$ q = \dfrac{{({a_1}{a_2} + {b_1}{b_2})}}{{{a^2}_2 + {b^2}_2}} + \dfrac{{({b_1}{a_2} - {b_2}{a_1})}}{{{a^2}_2 + {b^2}_2}}i $”.

Note: Complex number is a number that can be expressed in the form of $ a + ib $ , where $ a $ and $ b $ are real numbers, and $ i $ represents the imaginary unit, satisfying the equation $ {i^2} = - 1 $ . Because no real number satisfies this equation, $ i $ is called an imaginary number. Complex numbers allow solutions to certain equations that have no solutions in real numbers. The idea is to extend the real numbers with an intermediate $ i $ which is also called an imaginary unit taken to satisfy the relation $ {i^2} = - 1 $ , so that solutions to equations like the preceding one can be found.