Question

Let $z\in C$ be such that $\left| z \right|<1$. If $\omega =\dfrac{5+3z}{5\left( 1-z \right)}$ then :
A. $5\operatorname{Im}\left( \omega \right)<1$
B. $4\operatorname{Im}\left( \omega \right)<5$
C. $5\operatorname{Re}\left( \omega \right)>1$
D. $5\operatorname{Re}\left( \omega \right)>4$

Answer 1

Hint: Here in this question we have been given $\omega $ as a function of z in argand plane, so in this question, we have to first do cross multiplication, then take up z term to one side, use the property of modulus in complex numbers concept and then solve it to get the answer.

Complete step-by-step answer:
Here, in this question, we have been given, $\omega =\dfrac{5+3z}{5\left( 1-z \right)}$ where $z\in C$ and $\left| z \right|<1$. Each and every condition provided in this question is important for analysing the answer. So, we will first start by understanding the concept of complex numbers. A complex number is a number that can be expressed in the form of $a+ib$, where a and b are the real numbers and I is the imaginary unit, satisfying the equation ${{i}^{2}}=-1$. Since no number satisfies this equation, it is called the imaginary number. In the conceptual part, either the real or imaginary part can be zero. So, we can say that in $a+ib$, a is the real part and ib is the imaginary part. So, we will first simplify the equation given in the question by performing cross multiplication. So, we get,
$\begin{align}
  & \omega =\dfrac{5+3z}{5\left( 1-z \right)} \\
 & \Rightarrow 5\omega \left( 1-z \right)=5+3z \\
 & \Rightarrow 5\omega -5\omega z=5+3z \\
\end{align}$
Now, we will take the z terms to one side, so we get,
$5\omega -5=5\omega z+3z$
Taking out z common from the RHS, we get,
$5\omega -5=\left( 5\omega +3 \right)z$
Now, we apply modulus on both the sides. We know the property of modulus that if ${{z}_{1}}$ and ${{z}_{2}}$ are two complex numbers, then, $\left| {{z}_{1}}{{z}_{2}} \right|=\left| {{z}_{1}} \right|\left| {{z}_{2}} \right|$, so applying that we get,
$\begin{align}
  & \left| 5\omega -5 \right|=\left| \left( 5\omega +3 \right)z \right| \\
 & \Rightarrow \left| 5\omega -5 \right|=\left| 5\omega +3 \right|\left| z \right| \\
\end{align}$
Now, as we have been given that $\left| z \right|<1$, we can write as,
$\begin{align}
  & \left| 5\omega +3 \right|>\left| 5\omega -5 \right| \\
 & \Rightarrow \left| \dfrac{3}{5}+\omega \right|>\left| \omega -1 \right| \\
\end{align}$
Let us consider $\omega =x+iy$, then we get,
$\left| \dfrac{3}{5}+\left( x+iy \right) \right|>\left| \left( x+iy \right)-1 \right|$
If $z=x+iy$ is a complex number, then, $\left| z \right|=\left| x+iy \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}$. Expressing the above terms in the form $x+iy$, we get,
$\left| \left( \dfrac{3}{5}+x \right)+iy \right|>\left| \left( x-1 \right)+iy \right|$
Applying modulus, we get,
$\sqrt{{{\left( \dfrac{3}{5}+x \right)}^{2}}+{{y}^{2}}}>\sqrt{{{\left( x-1 \right)}^{2}}+{{y}^{2}}}$
Taking square on both sides, we get,
$\begin{align}
  & {{\left( \dfrac{3}{5}+x \right)}^{2}}+{{y}^{2}}>{{\left( x-1 \right)}^{2}}+{{y}^{2}} \\
 & \Rightarrow {{x}^{2}}+\dfrac{9}{25}+\dfrac{6}{5}x>{{x}^{2}}+1-2x \\
 & \Rightarrow \dfrac{16}{5}x>\dfrac{16}{25} \\
 & \Rightarrow x>\dfrac{1}{5} \\
 & \Rightarrow 5\operatorname{Re}\left( \omega \right)>1 \\
\end{align}$

So, the correct answer is “Option C”.

Note: The solution provided above is the shortest and best method to solve this question. But alternatively, we can consider $z=x+iy$ and then solve by expanding the same. Here, we might get confused by considering $\omega $ as the cube root of the unit but in this particular question, it just acts as a variable.