$|2z+3|=|z+6|$ where $z$ is a complex number represents a circle with center $(0,0)$ and radius $3$. It is fairly easy to prove it by letting $z=x+iy$ and we get the standard equation of circle in cartesian form. But is there any way we can deduce $|z|=3$ without exploiting $z=x+iy$ e.g in cartesian form? The reason being the cartesian calculation is a tad bit tedious. So it will be helpful if someone suggests a clever way of manipulating this.
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2 Answers
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Instead of working with the modulus directly, you can start with the two expressions $2z+3$ and $z+6$ and multiply each by their complex conjugate. Then you get:
$$4|z^2| + 12|z| +9 = |z^2| +12 |z| + 36$$
Simplifying leads to: $$|z^2| = 9$$
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$$|2z+3|=|z+6| \implies |2z+3|^2=|z+6|^2$$ $$ \implies (2z+3)(2z+3)^*=(z+6)(z+6)^* \implies (2z+3)(2z^*+3)=(z+6)(z^*+6)$$ $$\implies 4zz^*+6(z+z^*)+9 = zz^* + 6(z+z^*) + 36$$ $$\implies 3zz^* = 27 \implies 3|z|^2 = 27\implies |z|=3$$
