Theory of Intelligence
11TH•Mathematics•Complex Numbers•Question 12

Solution

If $z=x+iy$ and $\arg(z-2-3i)-\arg(z+2+3i)=2\pi$, show that $2y=3x$.

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Solution

Let z=x+iyz=x+iy.

z−2−3i=(x−2)+i(y−3),z+2+3i=(x+2)+i(y+3)z-2-3i=(x-2)+i(y-3),\quad z+2+3i=(x+2)+i(y+3)

Given:

arg⁡(z−2−3i)−arg⁡(z+2+3i)=2π\arg(z-2-3i)-\arg(z+2+3i)=2\pi

So:

arg⁡(z−2−3iz+2+3i)=2π\arg\left(\frac{z-2-3i}{z+2+3i}\right)=2\pi

That means the quotient lies on the positive real axis (argument 00 modulo 2π2\pi). Hence its imaginary part is 00.

Let:

w=(x−2)+i(y−3)(x+2)+i(y+3)w=\frac{(x-2)+i(y-3)}{(x+2)+i(y+3)}

Rationalize:

w=[(x−2)+i(y−3)] [(x+2)−i(y+3)](x+2)2+(y+3)2\begin{aligned} w&=\frac{[(x-2)+i(y-3)]\,[(x+2)-i(y+3)]}{(x+2)^2+(y+3)^2} \end{aligned}

Imaginary part of the numerator is:

(y−3)(x+2)−(x−2)(y+3)=4y−6x(y-3)(x+2)-(x-2)(y+3)=4y-6x

For ww to be real:

4y−6x=0⇒2y=3x4y-6x=0\Rightarrow 2y=3x 2y=3x\boxed{2y=3x}

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