Let arg(z) represent the principal argument o | vedclass

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(C) The equation $|z|=3$ represents a circle centered at the origin $(0,0)$ with radius $R=3$.The equation $\arg(z-1)-\arg(z+1)=\frac{\pi}{4}$ can be

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(C) The equation $|z|=3$ represents a circle centered at the origin $(0,0)$ with radius $R=3$.The equation $\arg(z-1)-\arg(z+1)=\frac{\pi}{4}$ can be rewritten as $\arg\left(\frac{z-1}{z+1}\right)=\frac{\pi}{4}$.This represents an arc of a circle passing through $z=1$ and $z=-1$.Let $z=x+iy$. The condition $\arg\left(\frac{z-1}{z+1}\right)=\frac{\pi}{4}$ implies that the locus is a circular arc with endpoints $(-1,0)$ and $(1,0)$.The center of this circle is $(0,1)$ and its radius is $\sqrt{2}$.The equation of this circle is $x^2+(y-1)^2=2$,which simplifies to $x^2+y^2-2y-1=0$.We need to find the intersection of $x^2+y^2=9$ and $x^2+y^2-2y-1=0$.Substituting $x^2+y^2=9$ into the second equation: $9-2y-1=0$,which gives $8-2y=0$,so $y=4$.However,for the circle $x^2+(y-1)^2=2$,the maximum value of $y$ is $1+\sqrt{2} \approx 2.414$.Since $4 > 1+\sqrt{2}$,the circle $|z|=3$ and the arc do not intersect.Therefore,they intersect nowhere.

Let $\arg(z)$ represent the principal argument of the complex number $z$. The curves $|z|=3$ and $\arg(z-1)-\arg(z+1)=\frac{\pi}{4}$ intersect:

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