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If $|z-1|=1$, find $\arg(z)$

Putting $z=r(\cos\theta+i\sin\theta)$ in $|z-1|=1$,

$$|r\cos\theta-1+ir\sin\theta|=1\\\implies r^2\cos^2\theta+1-2r\cos\theta+r^2\sin^2\theta=1\\\implies r=2\cos\theta$$

Putting $r$ in $z$, $$z=2\cos^2\theta+i2\cos\theta\sin\theta\\\implies|z-1|=\cos2\theta+i\sin2\theta$$

So, $\arg(z)=\frac12\arg(z-1)$, which matches with the answer but I am looking for any other method to do it. Maybe a more intuitive method or one involving a diagram?

aarbee
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2 Answers2

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The set of the points with $|z-1|=1$ is a circle around $1+0i$ with radius $1$. The argument of $z$ is the angle the vector has with the real axis. The answer $\arg(z)=\frac12\arg(z-1)$ corresponds to the familiar theorem regarding the inscribed angle.

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the vector $z_2 = z_1 - (1,0)$. To find the argument of $z_2$, you should translate it to the origin and calculate its angle with the $x$ axis. But this is the same as calculating the angle $\angle ABC$, which is why we can think of $\arg(z-1)$ as the argument of the point $z$ calculated from the center $1$.

Dosidis
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Yes, there is. Note that $|z-1|=1$ is equivalent to the circle $(x-1)^2+y^2=1$ in the Cartesian plane. This represents a circle with centre $C(1,0)$ and radius $1$. Now, take a random point $P$ on this circle. Let $O$ be the origin. Note that $\Delta OPC$ is an isosceles triangle with $OC=PC$. This means that $\angle POC=\angle OPC=\frac 12 \angle PCX$ from the exterior angle property of of triangle, where $\angle PCX$ represents angle of $PC$ with positive $x$-axis. This is equivalent to $arg(z)=\frac 12 arg(z-1)$ in the Argand plane, since $\arg(z-1)=\angle PCX$ and $arg(z)=\angle POC$.

Ritam_Dasgupta
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  • Hi. Is this only for circle that the argument angle is considered from the center of the circle? In general isn't argument take from origin? – aarbee Jul 05 '21 at 16:53
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    Indeed it is. The angle from the centre of the circle is $arg(z-1)$. You can consider $z$ to be a vector. Then, $1$, $z-1$ and $z$ vectors form a triangle as $z=(z-1)+1$. This means that $z-1$ is the vector joining the centre of the circle to the point $z$. Hence, $arg(z-1)$ is the angle that vector makes with the positive $x$ axis. – Ritam_Dasgupta Jul 05 '21 at 19:15