How to understand $|1+z|<1$ where $z\in\mathbb{C}$ geometrically?

Question

I am trying to understand how to plot $|1+z|<1$ where $z\in\mathbb{C}$, is it a circle centred at real axis $\text{Re}(z)=-1$ with radius $1$?

My question is how can we show it algebraically?

I tried $|1+z|=|1+a+bi|<1$, then I am not sure how to proceed to make it into equation of a circle?

Another question is what if $|1+z+\frac{z^2}{2}|<1$?

I tried $\Big|1+a+bi+\frac{(a+bi)^2}{2}\Big|=\Big|[1+a+\frac{1}{2}(a^2-b^2)]^2+(b+ab)^2 \Big|<1$? I am not sure whether it is a circle. Any hint?

Could somebody please give some help?

Many thanks!

Answer 1

Hint:

For $z=x+iy$ you have: $$ |1+x+iy|<1 $$ by definition of modulus for a complex number this means: $$ \sqrt{(1+x)^2+y^2}<1 $$

squaring.......

And you can do the same also for the other question.

Answer 2

Hint: For the first, write it as $$|z-(-1)|<1$$ Then you can see it is the set of points less than $1$ unit away from $-1$.

EDIT: Hmm, I see this does not really answer the question. Striking it out since it is irrelevant.

Answer 3

in the first problem |1+z|<1 take the argand plane relative to the point (-1,0) or -1+0i in the argand plane and thus 1+z becomes k(say). now plot point of all complex numbers relative to -1+0i in the imaginary plane with modulus 1 and you get a circle centered at (-1,0). thus all the points within the circle(|k|less than 1) will satisfy the equatins

Answer 4

I have no idea why you need an algebraic argument. $|1+z|<1$ means that the distance of the point $z$ from the point $-1$ is less than 1, so $z$ must lie in the open disk centre $-1$ radius 1. But if you put $z=a+ib$, then $|z+1|<1$ is equivalent to $|z+1|^2<1$ is equivalent to $(a+1)^2+b^2<1$ which is the open disk radius $1$, centred at $-1$.

For the second part you probably do need algebra. The same approach gives $(1+a+(a^2-b^2)/2)^2+(b+ab)^2<1$, which gives the interior of an oval around the point $-1$.