Rewriting $|z+2i|=\sqrt{2}|z+1|$ in $x+yi$ form with $x,y\in{}\mathbb{R}$

Question

According to my prof, the above equation is equivalent to $$x^2+(y+2)^2=2((x+1)^2+y^2)$$ I don't see how he could come to this conclusion, any insight/help would be highly appreciated.

Edit: It's probably very basic and I feel embarrassed asking such a question on here but still, am desperate on to knowing how.

Answer 1

\begin{align}\lvert z+2i\rvert=\sqrt2\lvert z+1\rvert&\iff\bigl\lvert x+(y+2)i)\bigr\rvert=\sqrt2\bigl\lvert x+1+yi\bigr\rvert\\&\iff\bigl\lvert x+(y+2)i)\bigr\rvert^2=2\bigl\lvert x+1+yi\bigr\rvert^2.\end{align}Can you take it from here?

Answer 2

If $z=x+iy$ then $|z|=\sqrt{x^2+y^2}$

Also, $z+2i=x+i(y+2)$ which leads to $|z+2i|=\sqrt{x^2+(y+2)^2}$

And, $z+1=(x+1)+iy$, which similarly leads to $\sqrt 2|z+1|=\sqrt 2 \sqrt{(x+1)^2+y^2}$

Try equating these then squaring them.