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This question already has an answer here:

Given that there are two complex numbers - $z, w$ - such that $w\overline{w} = 1$ and $z = \frac{1+w}{1-w}$, how do I deduce that $z$ lies on the imaginary axis?

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marked as duplicate by Arnaud D., Strants, Lord Shark the Unknown, max_zorn, José Carlos Santos Aug 29 '18 at 18:01

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  • $\begingroup$ It ends up being $i\dfrac1{\tan(\theta/2)}$, where $\theta$ is the argument of $w$ (the angle it makes with the real axis). Note that $w$ is on the unit circle. $\endgroup$ – Akiva Weinberger Aug 27 '15 at 20:15
  • $\begingroup$ How do you prove $w\overline{w} = 1$ $\endgroup$ – Lanous Feb 21 '17 at 3:21
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The obvious approach is to compute the conjugate. If $\bar z = -z$ then it is pure imaginary.

$$\bar z=\frac{1+\bar w}{1-\bar w}=\frac{w+w\bar w}{w-w\bar w}=\frac{w+1}{w-1}=-z$$

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The argument of $z = (1+w)/(1-w)$ is $\arg(1+w) - \arg(1-w)$. $z$ will be imaginary if and only if $\arg z = \pm \pi/2$.

The picture shows a vector with dashed line representing the complex number $w$, and two vectors with solid line representing $1+w$ and $1-w$.

The line from $1+w$ to $1-w$ is a diameter of the circle with unit radius and center $1$. Therefore the triangle defined by $0$, $1+w$ and $1-w$ has a right angle at $0$.

The figure also illustrates that the right angle at $0$ is the sum of $\arg(1+w)$ and $-\arg(1-w)$. Thus $\arg z = \arg(1+w) - \arg(1-w) = \pm \pi/2$ (it will be $\pi/2$ for $\arg w > 0$ as depicted in the figure, and $-\pi/2$ for $\arg w <0$). Therefore $z$ is imaginary.

enter image description here

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  • $\begingroup$ I prefer to read this proof the other way around, using the algebraic proof to verify that a triangle whose hypotenuse is the diameter of the circumcircle is a right triangle. $\endgroup$ – Mario Carneiro Aug 31 '15 at 18:39
  • $\begingroup$ @MarioCarneiro Your approach is neater :-) To add some variety, I wanted to offer a geometric argument. The fact that {a triangle with one side given by the diameter of the circle is a right triangle} is well known anyway $\endgroup$ – Luis Mendo Aug 31 '15 at 18:42

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