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Question: If z be a complex number \(\left| {z + 3} \right| \leqslant 8\). Then the value of \(\left| {z - 2} ...

If z be a complex number z+38\left| {z + 3} \right| \leqslant 8. Then the value of z2\left| {z - 2} \right| lies in
A)[2,13]A)[ - 2,13]
B)[0,13]B)[0,13]
C)[2,13]C)[2,13]
D)[13,2]D)[ - 13,2]

Explanation

Solution

First, complex numbers are the real and imaginary combined numbers as in the form of z=x+iyz = x + iy, where x and y are the real numbers and iiis the imaginary.
Imaginary ii can be also represented into the real values only if, i2=1{i^2} = - 1
The range of the function is given by z+38\left| {z + 3} \right| \leqslant 8 which means the value of z+3z + 3 has at most eight in positive or negative (because of the modulus).

Complete step-by-step solution:
From the given that zz is the complex number and z+38\left| {z + 3} \right| \leqslant 8.
The modulus value z+38\left| {z + 3} \right| \leqslant 8 can be rewritten as z+388z+38\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 x11x1\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 by3 - 3, thus we get only the complex number z in the center of the bounded function, hence we get, 8z+3811z5 - 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 z2\left| {z - 2} \right|, so we will again subtract the bounded values into two, thus we get, 13z23 - 13 \leqslant \left| {z - 2} \right| \leqslant 3
Hence the range of the function z2\left| {z - 2} \right| is minus thirteen to three.
Now we need to find z2\left| {z - 2} \right| lying points, hence z2[0,13]\left| {z - 2} \right| \in [0,13] because it will attain the value of the bounded function from 1313 to 00and that will be taken as the absolute value for the required answer.
Hence option B)[0,13]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]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 1313 to 00into the function z2\left| {z - 2} \right|.
Like apply the value of z as one, then we get, 12=1\left| {1 - 2} \right| = 1 which is in the range.
Again, apply the number eleven, then we get, 112=9\left| {11 - 2} \right| = 9 which is also in the range of z2\left| {z - 2} \right|.
Hence the range of the function must be zero to thirteen and it may be positive or negative.