Question

If $z=x+iy$ satisfies $\arg (z-1)=\arg (z+3i),$ then value of $\left( x-1 \right):y$ is equal to
$\begin{align}
  & \text{A}\text{. 2:1} \\
 & \text{B}\text{. 1:3} \\
 & \text{C}\text{. -1:3} \\
 & \text{D}\text{. None of these} \\
\end{align}$

Answer 1

Hint: First we substitute the value of $z$ in the given expression $\arg (z-1)=\arg (z+3i),$ and simplify the expression. Then use the property $\arg \left( a+ib \right)={{\tan }^{-1}}\dfrac{b}{a}$ and substitute the value. When we simplify the equations we get the value of $\left( x-1 \right):y$.

Complete step by step answer:
We have given that $z=x+iy$ satisfies $\arg (z-1)=\arg (z+3i),$
We have to find the value of $\left( x-1 \right):y$
Now, as given $z=x+iy$, when substitute the value in the given arg, we get
$\begin{align}
  & \arg (z-1)=\arg (z+3i) \\
 & \Rightarrow \arg (x+iy-1)=\arg (x+iy+3i) \\
 & \Rightarrow \arg \left( \left( x-1 \right)+iy \right)=\arg \left( x+i\left( y+3 \right) \right) \\
\end{align}$
Now, we know that $\arg \left( a+ib \right)={{\tan }^{-1}}\dfrac{b}{a}$
So, the equation becomes
$\Rightarrow {{\tan }^{-1}}\dfrac{y}{x-1}={{\tan }^{-1}}\dfrac{y+3}{x}$
When, we simplify the above equation we get
$\Rightarrow \dfrac{y}{x-1}=\dfrac{y+3}{x}$
Now, cross multiply the equations, we get
$\begin{align}
  & \Rightarrow xy=\left( y+3 \right)\left( x-1 \right) \\
 & \Rightarrow xy=xy-y+3x-3 \\
 & \Rightarrow xy-xy=-y+3x-3 \\
 & \Rightarrow 0=-y+3x-3 \\
 & \Rightarrow y=3\left( x-1 \right) \\
\end{align}$
Now, we have to find the value of $\left( x-1 \right):y$, so from above equation we get
$\dfrac{\left( x-1 \right)}{y}=\dfrac{1}{3}$

So, the correct answer is “Option B”.

Note: Alternatively students try to solve the question directly by simplifying the equations as
\[\begin{align}
  & \arg (z-1)=\arg (z+3i) \\
 & \Rightarrow \arg (x+iy-1)=\arg (x+iy+3i) \\
 & \Rightarrow \arg \left( \left( x-1 \right)+iy \right)=\arg \left( x+i\left( y+3 \right) \right) \\
 & \Rightarrow \left( \left( x-1 \right)+iy \right)=\left( x+i\left( y+3 \right) \right) \\
 & \Rightarrow \left( \left( x-1 \right)+iy \right)=\left( x+iy+3i \right) \\
 & \Rightarrow x-1+iy=x+iy+3i \\
 & \Rightarrow x-1=x+3i \\
\end{align}\]
but didn’t reach the conclusion. So, it is necessary to apply the property $\arg \left( a+ib \right)={{\tan }^{-1}}\dfrac{b}{a}$.