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Question: If \(z=x+iy\) satisfies \(\arg (z-1)=\arg (z+3i),\) then value of \(\left( x-1 \right):y\) is equal ...

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

Explanation

Solution

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

Complete step by step answer:
We have given that z=x+iyz=x+iy satisfies arg⁡(z−1)=arg⁡(z+3i),\arg (z-1)=\arg (z+3i),
We have to find the value of (x−1):y\left( x-1 \right):y
Now, as given z=x+iyz=x+iy, when substitute the value in the given arg, we get
arg⁡(z−1)=arg⁡(z+3i) ⇒arg⁡(x+iy−1)=arg⁡(x+iy+3i) ⇒arg⁡((x−1)+iy)=arg⁡(x+i(y+3)) \begin{aligned} & \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{aligned}
Now, we know that arg⁡(a+ib)=tan⁡−1ba\arg \left( a+ib \right)={{\tan }^{-1}}\dfrac{b}{a}
So, the equation becomes
⇒tan⁡−1yx−1=tan⁡−1y+3x\Rightarrow {{\tan }^{-1}}\dfrac{y}{x-1}={{\tan }^{-1}}\dfrac{y+3}{x}
When, we simplify the above equation we get
⇒yx−1=y+3x\Rightarrow \dfrac{y}{x-1}=\dfrac{y+3}{x}
Now, cross multiply the equations, we get
⇒xy=(y+3)(x−1) ⇒xy=xy−y+3x−3 ⇒xy−xy=−y+3x−3 ⇒0=−y+3x−3 ⇒y=3(x−1) \begin{aligned} & \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{aligned}
Now, we have to find the value of (x−1):y\left( x-1 \right):y, so from above equation we get
(x−1)y=13\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

& \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{aligned}$$ but didn’t reach the conclusion. So, it is necessary to apply the property $\arg \left( a+ib \right)={{\tan }^{-1}}\dfrac{b}{a}$.