Question

The modulus of \[\dfrac{{{\text{1 - i}}}}{{{\text{3 + i}}}}{\text{ + }}\dfrac{{{\text{4i}}}}{{\text{5}}}\]
A.\[\sqrt {\text{5}} \] unit.
B.\[\dfrac{{\sqrt {{\text{11}}} }}{{\text{5}}}\]unit
C.\[\dfrac{{\sqrt {\text{5}} }}{{\text{5}}}\] unit
D.\[\dfrac{{\sqrt {{\text{12}}} }}{{\text{5}}}\]unit

Answer 1

Hint: First, convert the given question in the form of \[z = x + iy\] by rationalizing the denominator. And then find
\[\left| {\text{z}} \right|{\text{ = }}\sqrt {\left( {{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}} \right)} \]
Rationalizing term is \[{\text{x + iy}}\] will be \[{\text{x - iy}}\]

Complete step-by-step answer:
Rationalizing the denominator in the given question,
\[{\text{ = }}\dfrac{{\left( {{\text{1 - i}}} \right)}}{{\left( {{\text{3 + i}}} \right)}}{{ \times }}\dfrac{{\left( {{\text{3 - i}}} \right)}}{{\left( {{\text{3 - i}}} \right)}}{\text{ + }}\dfrac{{{\text{4i}}}}{{\text{5}}}\]
\[{\text{ = }}\dfrac{{\left( {{\text{1 - i}}} \right)\left( {{\text{3 - i}}} \right)}}{{\left( {{\text{3 + i}}} \right)\left( {{\text{3 - i}}} \right)}}{\text{ + }}\dfrac{{{\text{4i}}}}{{\text{5}}}\]
Now, $\left( {{\text{a + b}}} \right)\left( {{\text{a - b}}} \right){\text{ = }}{{\text{a}}^{\text{2}}}{\text{ - }}{{\text{b}}^{\text{2}}}$
\[{\text{ = }}\dfrac{{\left( {{{\text{i}}^{\text{2}}}{\text{ - 3i - i + 3}}} \right)}}{{{{\text{3}}^{\text{2}}}{\text{ - }}{{\text{i}}^{\text{2}}}}}{\text{ + }}\dfrac{{{\text{4i}}}}{{\text{5}}}\]
\[{\text{ = }}\dfrac{{{\text{ - 1 - 3i - i + 3}}}}{{{\text{9 - }}\left( {{\text{ - 1}}} \right)}}{\text{ + }}\dfrac{{{\text{4i}}}}{{\text{5}}}\]
\[{\text{ = }}\dfrac{{{\text{2 - 4i}}}}{{{\text{10}}}}{\text{ + }}\dfrac{{{\text{4i}}}}{{\text{5}}}\]
\[{\text{ = }}\dfrac{{{\text{1 - 2i}}}}{{\text{5}}}{\text{ + }}\dfrac{{{\text{4i}}}}{{\text{5}}}\]
\[{\text{ = }}\dfrac{{{\text{1 + 2i}}}}{{\text{5}}}\]
Hence, \[{\text{z = }}\dfrac{{{\text{1 + 2i}}}}{{\text{5}}}\]
Compare the above, with \[{\text{z = x + iy}}\]
${\text{x = }}\dfrac{{\text{1}}}{{\text{5}}}$ and ${\text{y = }}\dfrac{{\text{2}}}{{\text{5}}}$
Now, using \[\left| {\text{z}} \right|{\text{ = }}\sqrt {\left( {{{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}} \right)} \]
\[\left| {\text{z}} \right|{\text{ = }}\sqrt {\left( {{{\left( {\dfrac{{\text{1}}}{{\text{5}}}} \right)}^{\text{2}}}{\text{ + }}{{\left( {\dfrac{{\text{2}}}{{\text{5}}}} \right)}^{\text{2}}}} \right)} \]
\[\left| {\text{z}} \right|{\text{ = }}\sqrt {\left( {\dfrac{{\text{1}}}{{{\text{25}}}}{\text{ + }}\dfrac{{\text{4}}}{{{\text{25}}}}} \right)} {\text{ = }}\sqrt {\dfrac{{\text{5}}}{{{\text{25}}}}} \]
\[\left| {\text{z}} \right|{\text{ = }}\dfrac{{\sqrt {\text{5}} }}{{\text{5}}}\]
Hence option (c) is the correct answer.

Note: Remember the general notation of complex numbers and remember the formula to calculate the modulus of complex numbers. Also convert the denominator into rational and real integral form in order to convert it into standard formulas.
A rational number is a number that can be expressed as the ratio of two integers. ... However, one third can be expressed as 1 divided by 3, and since 1 and 3 are both integers, one third is a rational number. Likewise, any integer can be expressed as the ratio of two integers, thus all integers are rational.