How do you simplify \[\dfrac{{4 + 2i}}{{4 - 2i}}\] ?
Answer
611.7k+ views
Hint: The problem is of real and imaginary numbers. We will first find the complex conjugate of denominators. Then we will multiply and divide both numerator and denominator by this complex conjugate. Then this will simplify the denominator of the number such that the denominator is totally a real number.
Complete step-by-step solution:
Given ratio is a ratio of complex numbers. In order to solve this we will take the complex conjugate of denominators. Denominator \[4 - 2i\] has complex conjugate \[4 + 2i\].
Now we will multiply numerator and denominator by complex conjugate.
\[ \Rightarrow \dfrac{{4 + 2i}}{{4 - 2i}} \times \dfrac{{4 + 2i}}{{4 + 2i}}\]
Now the denominator is in the form \[\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}\]
\[ \Rightarrow \dfrac{{\left( {4 + 2i} \right)\left( {4 + 2i} \right)}}{{{4^2} - {{\left( {2i} \right)}^2}}}\]
Now taking squares of terms in denominator and multiplying the terms in numerator,
\[ \Rightarrow \dfrac{{4 \times 4 + 4 \times 2i + 2i \times 4 + {{\left( {2i} \right)}^2}}}{{16 - 4{i^2}}}\]
Solving the terms in numerator,
\[ \Rightarrow \dfrac{{16 + 8i + 8i + 4{i^2}}}{{16 - 4{i^2}}}\]
Adding the complex numbers we get,
\[ \Rightarrow \dfrac{{16 + 16i + 4{i^2}}}{{16 - 4{i^2}}}\]
Taking 4 common from numerator and denominator and cancelling it,
\[ \Rightarrow \dfrac{{4 + 4i + {i^2}}}{{4 - {i^2}}}\]
Now we know that \[{i^2} = - 1\] so substituting it we get,
\[ \Rightarrow \dfrac{{4 + 4i - 1}}{{4 - \left( { - 1} \right)}}\]
On rearranging the terms,
\[ \Rightarrow \dfrac{{4 - 1 + 4i}}{{4 + 1}}\]
Adding the terms in denominator and in numerator,
\[ \Rightarrow \dfrac{{3 + 4i}}{5}\]
Separating the real and imaginary part we get,
\[ \Rightarrow \dfrac{3}{5} + \dfrac{4}{5}i\]
Therefore the correct answer is \[\dfrac{3}{5} + \dfrac{4}{5}i\]
Note: These types of problems generally proceed with complex conjugate. Complex conjugate is the same number as original such that the real part is the same and the imaginary part is only different in sign. \[z = x + iy\] is the complex number with x as real part and y as imaginary part. Don’t forget to place the value of \[{i^2} = - 1\].
Complete step-by-step solution:
Given ratio is a ratio of complex numbers. In order to solve this we will take the complex conjugate of denominators. Denominator \[4 - 2i\] has complex conjugate \[4 + 2i\].
Now we will multiply numerator and denominator by complex conjugate.
\[ \Rightarrow \dfrac{{4 + 2i}}{{4 - 2i}} \times \dfrac{{4 + 2i}}{{4 + 2i}}\]
Now the denominator is in the form \[\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}\]
\[ \Rightarrow \dfrac{{\left( {4 + 2i} \right)\left( {4 + 2i} \right)}}{{{4^2} - {{\left( {2i} \right)}^2}}}\]
Now taking squares of terms in denominator and multiplying the terms in numerator,
\[ \Rightarrow \dfrac{{4 \times 4 + 4 \times 2i + 2i \times 4 + {{\left( {2i} \right)}^2}}}{{16 - 4{i^2}}}\]
Solving the terms in numerator,
\[ \Rightarrow \dfrac{{16 + 8i + 8i + 4{i^2}}}{{16 - 4{i^2}}}\]
Adding the complex numbers we get,
\[ \Rightarrow \dfrac{{16 + 16i + 4{i^2}}}{{16 - 4{i^2}}}\]
Taking 4 common from numerator and denominator and cancelling it,
\[ \Rightarrow \dfrac{{4 + 4i + {i^2}}}{{4 - {i^2}}}\]
Now we know that \[{i^2} = - 1\] so substituting it we get,
\[ \Rightarrow \dfrac{{4 + 4i - 1}}{{4 - \left( { - 1} \right)}}\]
On rearranging the terms,
\[ \Rightarrow \dfrac{{4 - 1 + 4i}}{{4 + 1}}\]
Adding the terms in denominator and in numerator,
\[ \Rightarrow \dfrac{{3 + 4i}}{5}\]
Separating the real and imaginary part we get,
\[ \Rightarrow \dfrac{3}{5} + \dfrac{4}{5}i\]
Therefore the correct answer is \[\dfrac{3}{5} + \dfrac{4}{5}i\]
Note: These types of problems generally proceed with complex conjugate. Complex conjugate is the same number as original such that the real part is the same and the imaginary part is only different in sign. \[z = x + iy\] is the complex number with x as real part and y as imaginary part. Don’t forget to place the value of \[{i^2} = - 1\].
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