Determine the set of all complex numbers satisfying the relation $|z|<4$, here $z$ is a complex number.

Question

Please, solve this above question. Please, tell me how to find domain of a function in terms of complex numbers. I can find domain in terms of real no, but facing issue with complex ones.

If I consider $z=a+bi$, $a,b \in \mathbb R$. then I can get $a^2+b^2<16$. so the real numbers should the interior points of the circle $a^2+b^2=16$. But, what about complex ?

Answer 1

What you showed is the solution. You can identify each complex number $z = a + bi$ with a point $(a, b)$ in the plane $\mathbb{R}^{2}$. This means that $\mathbb{C}$ and $\mathbb{R}^{2}$ are isomorphic.

The complex modulus (absolute value) is defined to be

$$|z| := \sqrt{Re(z)^{2} + Im(z)^{2}} = \sqrt{a^{2} + b^{2}},$$

where $Re(z) := a$ and $Im(z) := b$ are the real and imaginary parts of $z = a + bi$, respectively.

Thus, when you are asking $|z| < 4$, you are asking for $|z| = \sqrt{a^{2} + b^{2}} < 4$.

You can see a visualization of the complex number in the complex plane here:

complex number geometry

Answer 2

The points z such that $|z|<4$ are all the points that lie within the disk of radius 4 centered at 0. Let $z=a+ib$, $|z|<4$ implies $a^2+b^2<16$.