A complex number is such that the absolute value of z = the absolute value of (z-3i)
(a) Show that the imaginary part of z is 2/3.
I tried squaring both sides and then I find by equating both sides, that when z = a + bi, b satisfies the equation:
b^2 -7b +9 =0 however this does not give me b = 2/3.
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You have $$a^2+b^2=a^2+(b-3)^2\iff6b=9\iff2b=3$$
or $$b^2-(b-3)^2=0\iff\{b-(b-3)\}\{b+(b-3)\}=0$$
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Geometry! (In the complex plane...)
The distance $|z|=|z-0|$ between the points $z$ and $0$ and the distance $|z-3\mathrm i|$ between the points $z$ and $3\mathrm i$ coincide if and only if the point $z$ is on the line orthogonal to the segment $[0,3\mathrm i]$ and passing through its middle point $\frac12(0+3\mathrm i)=\frac32\mathrm i$. This is the horizontal line $\frac32\mathrm i+\mathbb R$ of equation $\Im(z)=\frac32$.
Let$$z_1=a+ib$$
$$z_2=z-3i=a+i(b-3)$$ By comparing Modulus of $z_1$ and $z_2$ $$|z_1|=|z_2|$$ $$a^2+b^2=a^2+(b-3)^2$$ $$a^2+b^2=a^2+b^2-6b+9$$
$$6b=9$$
$$b=\frac32$$