In Gaussian plane draw solution of equation |z-(1+2i)|=2
My solution:
Wolfram solution:
I don't understand, why my solution is not right. Could anyone help me, please?
In Gaussian plane draw solution of equation |z-(1+2i)|=2
My solution:
Wolfram solution:
I don't understand, why my solution is not right. Could anyone help me, please?
For some reason, W|A seems to be depicting $(1+2i)\pm2$. Now in the reals the solutions to the equation $|z-a|=b$ are $z=a\pm b$, but not in $\Bbb C$, so it might have been conflating it with that.
Your depiction of the solution set is perfectly correct.
$|z-(1+2i)|=2 \iff z-(1+2i)=2 e^{i\theta}$ for all $\theta \in\mathbb {R}$
$\iff z=1+2i+2e^{i\theta}$ for all $\theta\in\mathbb{R}$
When we take $\theta = 0, -\pi, -\frac{\pi}{2}$ and $\frac{\pi}{2}$,
we have $z=3+2i$, $-1+2i$, $1$ and $1+4i$.