Question

How do you find the absolute value of \[-2-i\]?

Answer 1

Hint: Now the given number is a complex number of the form $a+ib$ where $a=-2$ and $b=-1$ . Now we know that the absolute value of a complex number of the form $a+ib$ is given by $\sqrt{{{a}^{2}}+{{b}^{2}}}$ hence we will substitute the value of a and b and find the absolute value of the complex number.

Complete step by step solution:
Let us first understand the concept of complex numbers.
Now we know the real numbers which are either rational or irrational.
Now we know that square of a number is positive. Hence we cannot take the square root of negative numbers. Hence we do not have $\sqrt{-1}$ in real numbers.
Hence we define complex numbers. Now complex numbers are formed adding imaginary numbers in the set of real numbers.
Now the number $\sqrt{-1}$ is denoted by I and is called iota.
Now the general form of the complex number is $a+ib$ where a and b both are real.
Now absolute value of a complex number is given by $\sqrt{{{a}^{2}}+{{b}^{2}}}$ and is denoted by $\left| z \right|$ .
Now consider the given number \[-2-i\]
Comparing $a+ib$ we get a = - 2 and b = - 1.
Hence the absolute value of the number is given by $\sqrt{{{\left( -2 \right)}^{2}}+{{\left( -1 \right)}^{2}}}=\sqrt{4+1}=\sqrt{5}$
Hence the absolute value of the number is $\sqrt{5}$ .

Note: Now note that a real number is also a complex number as it can be written in the form $a+0i$ . Hence the set of real numbers is just a subset of complex numbers. Now note that the absolute value of a real number is just taking modulus of the function. While in complex numbers we take $\sqrt{{{a}^{2}}+{{b}^{2}}}$.