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Question: If \(\left| {z - \dfrac{4}{z}} \right| = 2\), then the maximum value of \(\left| z \right|\) is equa...

If z4z=2\left| {z - \dfrac{4}{z}} \right| = 2, then the maximum value of z\left| z \right| is equal to
A. 3+1 B. 5+1 C. 2 D. 2 + 2  {\text{A}}{\text{. }}\sqrt 3 + 1 \\\ {\text{B}}{\text{. }}\sqrt 5 + 1 \\\ {\text{C}}{\text{. 2}} \\\ {\text{D}}{\text{. 2 + }}\sqrt 2 \\\

Explanation

Solution

We have given z4z=2\left| {z - \dfrac{4}{z}} \right| = 2 and we have to find maximum value of z\left| z \right| so we have to start from given and use the concept of modulus and complex number so that we can make equation in form of z\left| z \right| and then we will be able to find maximum value of z\left| z \right|.

Complete step by step answer:
We have given
z4z=2\left| {z - \dfrac{4}{z}} \right| = 2
And we have to find z\left| z \right|
Now we can write z\left| z \right|= z4z+4z\left| {z - \dfrac{4}{z} + \dfrac{4}{z}} \right|
And we know the property of complex number
a+ba+b\left| {a + b} \right| \leqslant \left| a \right| + \left| b \right|
We will use this property in above equation as a=z4z,b=4za = z - \dfrac{4}{z},b = \dfrac{4}{z}
On applying this property we get,
zz4z+4z\left| z \right| \leqslant \left| {z - \dfrac{4}{z}} \right| + \left| {\dfrac{4}{z}} \right|
As we have given in question z4z=2\left| {z - \dfrac{4}{z}} \right| = 2 putting this value in above equation we get,
z2+4z\left| z \right| \leqslant 2 + \dfrac{4}{{\left| z \right|}}
On simplifying we get,
z24z0\left| z \right| - 2 - \dfrac{4}{{\left| z \right|}} \leqslant 0
z22z4z0\dfrac{{{{\left| z \right|}^2} - 2\left| z \right| - 4}}{{\left| z \right|}} \leqslant 0
We can write it as also
z22z4=0{\left| z \right|^2} - 2\left| z \right| - 4 = 0
Here this is quadratic equation in z\left| z \right|
And for the roots of quadratic equation we know the formula
If ax2+bx+c=0a{x^2} + bx + c = 0 is the form of quadratic equation then its roots are
x=b±b24ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}
Here applying same concept we get,
Hence roots of the quadratic equations are
z=±2+4+162 =1±5  z = \pm 2 + \dfrac{{ - \sqrt {4 + 16} }}{2} \\\ = 1 \pm \sqrt 5 \\\
And hence
0z1+50 \leqslant \left| z \right| \leqslant 1 + \sqrt 5
And so max z=1+5\left| z \right| = 1 + \sqrt 5
Hence option B is the correct option.

Note:
Whenever we get this type of question the key concept of solving is we have to must remember properties of complex number and modulus like a+ba+b\left| {a + b} \right| \leqslant \left| a \right| + \left| b \right|. These help in solving questions easily.