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Question

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

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

Explanation

Solution

Now the given number is a complex number of the form a+iba+ib where a=2a=-2 and b=1b=-1 . Now we know that the absolute value of a complex number of the form a+iba+ib is given by a2+b2\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 1\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 1\sqrt{-1} is denoted by I and is called iota.
Now the general form of the complex number is a+iba+ib where a and b both are real.
Now absolute value of a complex number is given by a2+b2\sqrt{{{a}^{2}}+{{b}^{2}}} and is denoted by z\left| z \right| .
Now consider the given number 2i-2-i
Comparing a+iba+ib we get a = - 2 and b = - 1.
Hence the absolute value of the number is given by (2)2+(1)2=4+1=5\sqrt{{{\left( -2 \right)}^{2}}+{{\left( -1 \right)}^{2}}}=\sqrt{4+1}=\sqrt{5}
Hence the absolute value of the number is 5\sqrt{5} .

Note: Now note that a real number is also a complex number as it can be written in the form a+0ia+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 a2+b2\sqrt{{{a}^{2}}+{{b}^{2}}}.