Question

Given $z$ is a complex number with modulus 1. Then the equation in a, ${\left( {\dfrac{{1 + ia}}{{1 - ia}}} \right)^4} = z$ has
A.All roots are real and distinct.
B.Two real and two imaginary.
C.Three roots real and one imaginary
D.One root real and three imaginary

Answer 1

Hint: In order to solve the given problem first we will use the concept of imaginary will modulus 1 in order to substitute some general value in the place of z. Further we will substitute the same in the problem equation to solve further and bring the equation in some general terms to find the type of root of the equation amongst the given options.

Complete step-by-step answer:
Given that: $z$ is a complex number with modulus 1
And also ${\left( {\dfrac{{1 + ia}}{{1 - ia}}} \right)^4} = z$
As we know that any imaginary number with modulus can be represented as a circle in the imaginary plane so the complex number z can be substituted with the general equation of a circle in the imaginary plane which is given by:
$z = cis\left( P \right) = \cos \left( P \right) + \sin \left( P \right)$
Now let us substitute the value of z from the above equation into the problem equation so we get:
${\left( {\dfrac{{1 + ia}}{{1 - ia}}} \right)^4} = z = cis\left( P \right) = \cos \left( P \right) + \sin \left( P \right) \\\ \Rightarrow \left( {\dfrac{{1 + ia}}{{1 - ia}}} \right) = {\left( z \right)^{\dfrac{1}{4}}} = {\left[ {cis\left( P \right)} \right]^{\dfrac{1}{4}}} \\\$
According to De-Moivre's theorem we know that the term ${\left[ {cis\left( P \right)} \right]^{\dfrac{1}{4}}}$ can be written in a sample form as follows:
${\left[ {cis\left( P \right)} \right]^{\dfrac{1}{4}}} = cis\dfrac{{2k\pi + P}}{4} \\\ {\text{where }}k = 0,1,2,3 \\\$
Let us consider the angle as follows
$Q = \dfrac{{2k\pi + P}}{4}$
So we have:
${\left[ {cis\left( P \right)} \right]^{\dfrac{1}{4}}} = cis\dfrac{{2k\pi + P}}{4} = cis\left( Q \right)$
Let us now substitute the general value and the considered value of the angle in the problem equation. So we obtain:
$\Rightarrow \left( {\dfrac{{1 + ia}}{{1 - ia}}} \right) = {\left[ {cis\left( P \right)} \right]^{\dfrac{1}{4}}} = cis\left( Q \right) \\\ \Rightarrow \left( {\dfrac{{1 + ia}}{{1 - ia}}} \right) = cis\left( Q \right) \\\$
Let us now manipulate the above equation in order to find the value of a.
$\Rightarrow 1 + ia = \left( {1 - ia} \right)cis\left( Q \right) \\\ \Rightarrow 1 + ia = cis\left( Q \right) - \left( {ia \times cis\left( Q \right)} \right) \\\ \Rightarrow ia + \left( {ia \times cis\left( Q \right)} \right) = cis\left( Q \right) - 1 \\\ \Rightarrow ia\left( {1 + cis\left( Q \right)} \right) = cis\left( Q \right) - 1 \\\ \Rightarrow ia = \dfrac{{ - 1 + cis\left( Q \right)}}{{1 + cis\left( Q \right)}} \\\$
Now, let us substitute the general value of $cis\left( Q \right)$
$\Rightarrow ia = \dfrac{{\sin \dfrac{Q}{2}\left( {i\cos \dfrac{Q}{2} - \sin \dfrac{Q}{2}} \right)}}{{\cos \dfrac{Q}{2}\left( {\cos \dfrac{Q}{2} + i\sin \dfrac{Q}{2}} \right)}}$
Further let us bring out the common terms from the numerator and denominator in order to simply the terms.
$\Rightarrow ia = \dfrac{{i\sin \dfrac{Q}{2}\left( {\cos \dfrac{Q}{2} + i\sin \dfrac{Q}{2}} \right)}}{{\cos \dfrac{Q}{2}\left( {\cos \dfrac{Q}{2} + i\sin \dfrac{Q}{2}} \right)}} \\\ \Rightarrow ia = \dfrac{{i\sin \dfrac{Q}{2}}}{{\cos \dfrac{Q}{2}}} \\\ \Rightarrow ia = i\tan \dfrac{Q}{2} \\\ \Rightarrow a = \tan \dfrac{Q}{2} \\\$
From the above solution equation as we can see the value of terms “a” is in terms of general value considered and in the form of tangent of angle.
So the roots of the equation are real and are distinct for different values of k.
Hence, all the roots are real and distinct.
So, option A is the correct option.

Note:In order to solve the problems related to complex numbers and their roots students must remember the general form of the complex number in view of the complex plane. Also for solving such problems a general complex term can be considered and further proceeded. Students must also remember the curve for different trigonometric terms.