Question

If z be a complex number $\left| {z + 3} \right| \leqslant 8$. Then the value of $\left| {z - 2} \right|$ lies in
$A)[ - 2,13]$
$B)[0,13]$
$C)[2,13]$
$D)[ - 13,2]$

Answer 1

Hint: First, complex numbers are the real and imaginary combined numbers as in the form of $z = x + iy$, where x and y are the real numbers and $i$is the imaginary.
Imaginary $i$ can be also represented into the real values only if, ${i^2} = - 1$
The range of the function is given by $\left| {z + 3} \right| \leqslant 8$ which means the value of $z + 3$ has at most eight in positive or negative (because of the modulus).

Complete step-by-step solution:
From the given that $z$ is the complex number and $\left| {z + 3} \right| \leqslant 8$.
The modulus value $\left| {z + 3} \right| \leqslant 8$ can be rewritten as $\left| {z + 3} \right| \leqslant 8 \Rightarrow - 8 \leqslant z + 3 \leqslant 8$ (because of the modulus value it can be positive or negative, the values will not be changed).
Since this can be obtained from the result of the complex number $\left| x \right| \leqslant 1 \Rightarrow - 1 \leqslant x \leqslant 1$ (will be the bounded values of the given function)
Now we are going the subtract by$ - 3$, thus we get only the complex number z in the center of the bounded function, hence we get, $ - 8 \leqslant z + 3 \leqslant 8 \Rightarrow - 11 \leqslant z \leqslant 5$ (by the subtraction operation)
Since the given question ask us to find the value of $\left| {z - 2} \right|$, so we will again subtract the bounded values into two, thus we get, $ - 13 \leqslant \left| {z - 2} \right| \leqslant 3$
Hence the range of the function $\left| {z - 2} \right|$ is minus thirteen to three.
Now we need to find $\left| {z - 2} \right|$ lying points, hence $\left| {z - 2} \right| \in [0,13]$ because it will attain the value of the bounded function from $13$ to $0$and that will be taken as the absolute value for the required answer.
Hence option $B)[0,13]$ is correct. (Since the complex number z is in the modulus so positive and negative values in the bounded functions are the same)
For option $C)[2,13]$ is also correct, but it will not contain the numbers zero and one.

Note: We can also able to check this by applying for the numbers from $13$ to $0$into the function $\left| {z - 2} \right|$.
Like apply the value of z as one, then we get, $\left| {1 - 2} \right| = 1$ which is in the range.
Again, apply the number eleven, then we get, $\left| {11 - 2} \right| = 9$ which is also in the range of $\left| {z - 2} \right|$.
Hence the range of the function must be zero to thirteen and it may be positive or negative.