Question

Find the modulus of the complex number $Z = 2 + 3i$.

Answer 1

Hint:Let $Z = a + bi$ be a complex number. Then, the modulus of a complex number $Z$ , denoted by $\left| Z \right|$ , is defined to be the non-negative real number $\left| Z \right| = \sqrt {{{\left( {\operatorname{Re} \left( z \right)} \right)}^2} + {{\left( {\operatorname{Im} \left( z \right)} \right)}^2}} = \sqrt {{a^2} + {b^2}}$.

Complete step-by-step answer:
Given, complex number $Z = 2 + 3i$ .
Real part of complex number $Z$ is $\operatorname{Re} \left( Z \right) = a = 2$ .
Imaginary part of complex number $Z$ is $\operatorname{Im} \left( Z \right) = b = 3$ .
Now, we apply the formula of modulus of complex number $Z$ .
$\left| Z \right| = \sqrt {{{\left( {\operatorname{Re} \left( z \right)} \right)}^2} + {{\left( {\operatorname{Im} \left( z \right)} \right)}^2}} = \sqrt {{a^2} + {b^2}}$
Put the value of a and b in the above formula.

$$\Rightarrow \left| Z \right| = \sqrt {{{\left( {\operatorname{Re} \left( z \right)} \right)}^2} + {{\left( {\operatorname{Im} \left( z \right)} \right)}^2}} = \sqrt {{a^2} + {b^2}} \\\ \Rightarrow \left| Z \right| = \sqrt {{{\left( 2 \right)}^2} + {{\left( 3 \right)}^2}} \\\ \Rightarrow \left| Z \right| = \sqrt {4 + 9} \\\ \Rightarrow \left| Z \right| = \sqrt {13} \\\$$

So, the modulus of complex number $Z = 2 + 3i$ is $\sqrt {13}$ .

Note: Whenever we face such types of problems we use some important points. First we find real and imaginary parts of complex numbers then apply the formula of modulus of complex number then after solving we can get the required answer.