Complex numbers (showing simplicity)

Algebra Level 3

If the complex number $z=x+iy$ satisfy the equation $\text{amp}(z-1)=\text{amp}(z+3)$ , then the value of $(x-1):y$ is equal to?

Clarification: $\text{amp}(z)$ means amplitude or argument of the complex number $z$ .

1:3 does not exist 2:1 -1:3

1 solution

Anandhu Raj
Feb 12, 2015

Let $z=x+iy$

Given $amp(z-1)=amp(z+3)$ $\Longrightarrow amp((x-1)+iy)=amp((x+3)+iy)$

$\Longrightarrow amp((x-1)+iy)-amp((x+3)+iy)=0$

$\Longrightarrow { tan }^{ -1 }\left| \frac { y }{ x-1 } \right| -{ tan }^{ -1 }\left| \frac { y }{ x+3 } \right| =0$

$\Longrightarrow { tan }^{ -1 }\left( \frac { \frac { y }{ x-1 } -\frac { y }{ x+3 } }{ 1+\frac { { y }^{ 2 } }{ (x-1)(x+3) } } \right) =0$

Solving gives,

$y=0$

Thus when we take ratio $(x-1):y$ $\longrightarrow$ $(x-1):0$ ,which is not defined.

But the difference of the inverse tangents can also be equal to pi

Abhishek Ranjan - 6 years, 3 months ago

Amplitude is not always $tan^-1(\frac{y}{x}$ but $\pi\pm tan^-1(\frac{y}{x}$ or $\pm tan^-1(\frac{y}{x}$ for 3,2 and 1,4 quadrant respectively.

Vishal Yadav - 4 years, 2 months ago

Its ok by checking one case, that is their differnece is 0. Its not a multicorrect question. So no tension

Md Zuhair - 3 years, 10 months ago