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Describe the locus of the following points on the Argand diagram:

$$\left|\frac{(z+1-i)}{(z-1-i)}\right| = 1$$and

$$\mathrm{arg}\left[\frac{(z+1+i)}{(z-1-i)}\right] = \pm \frac{\pi}{2}.$$

I would really love some help on these 2 questions, all responses much appreciated!

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What have you tried? Are you confused with some particular step? It helps for us to know. Regards –  Amzoti Apr 1 '13 at 4:12
    
I've tried putting z = x+iy then rationalising the denominator by multiplying by the conjugate which gave me some numbers, but I'm not sure what to do with them. –  Andy Apr 1 '13 at 4:22
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2 Answers

The first one is saying that the distance between $z$ and $1 + i$ is the same as the distance between $z$ and $1 - i$. The set of points equidistant from two points is the line bisecting the line segment joining the two points. Hence, the locus is the line $y = 0$.

The second one is saying if you look at the point $P$ given by translating $z$ 1 up and 1 to the right and the point $Q$ given by $z$ translated 1 down and 1 to the left, the angle between $P$ and $Q$ (with respect to the origin) is $\pi/2$. Intuitively, this seems to me to be the line $y = x$, though I haven't checked this rigorously.

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Ok thank you for the explanations, so for the first question it's simply the line x=0? Also, for the second question, I thought the angle was always taken between the positive direction of the horizontal - so from which point would we take the angle (does it matter)? –  Andy Apr 1 '13 at 4:25
    
Ooops, I meant $y = 0$, but yes, that's the answer. As for the second one, arg of a quotient is the angle between the numerator and the denominator. (Because multiplying two complex numbers results in the angles that they make with the positive horizontal axis being added) –  Jonah Sinick Apr 1 '13 at 4:35
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$$\left|\frac{(z+1-i)}{(z-1-i)}\right| = 1$$ $$\left|\ (z+1-i)\right| = |(z-1-i)|$$ put z=x+iy $$\left|\ (x+iy+1-i)\right| = |(x+iy-1-i)|$$ $$\left|\ (x+1)+i(y-1)\right| = |(x-1)+i(y-1)|$$ $$\left|\ (x+1)+i(y-1)\right|^2 = |(x-1)+i(y-1)|^2$$ since $$\left|\ (x+iy)\right|^2 = x^2+y^2$$ then $$\left|\ (x+1)+i(y-1)\right|^2 = |(x-1)+i(y-1)|^2$$ $$ (x+1)^2+(y-1)^2=(x-1)^2+(y-1)^2 $$ $$(x+1)^2=(x-1)^2$$ $$x=0$$ equation of Y-axis.

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