Simple complex numbers

$|a+bi|\le 5\quad \Rightarrow a^2+b^2 \le 25$ . The acceptable range of values of $a$ and $b$ are: $-5 \le a,b \le 5$ .

The following table gives $a^2+b^2$ with $a^2+b^2 \le 25$ shaded red. The number of acceptable $(a,b)$ pairs, and hence the acceptable number of complex numbers $a+bi$ is $\boxed{81}$ .

The question is essentially Gauss's Circle Problem for a radius of $r=5$ . If $N(r)$ is the number of complex numbers with integer components with magnitude not exceeding $r$ , then $N(r)=1+4[r]+\displaystyle \sum_{i=1}^{[r]} [\sqrt{r^2-i^2}]$ , where $[x]$ is $x$ rounded down. Thus, $N(5)=21+4\cdot ([\sqrt{24}]+[\sqrt{21}]+[\sqrt{16}]+[\sqrt{9}]+[\sqrt{0}])$ $=21+4(4+4+4+3+0)=21+60=\boxed{81}$ .

By the definition of absolute value of a complex number, $|a + b\text{i}| \equiv \sqrt{a^2 + b^2}.$ So to find all complex numbers with magnitude less than or equal to $5,$ use the inequality $a^2 + b^2 \leq 25.$ Since we only want the integer solutions like the problem asks, there are a finite number of solutions. So consider the positive solutions first... count them all, you will find there are $15$ positive-only solutions. Since the parity of $a^2$ is even, there will be the same number of solutions for each quadrant; that is, $15\times 4 = 60$ solutions per quadrant.

But we didn't count solutions where $a$ or $b$ is $0.$ Counting those, there will be $2\times 11 - 1$ solutions since we can't count $(0,0)$ twice. So the answer is $60 + 21 = \boxed{81}$

$\left| a+bi \right| \le 5\Leftrightarrow \sqrt { { a }^{ 2 }+{ b }^{ 2 } } \le 250$ . Hence, ${ a }^{ 2 }\le 25\Leftrightarrow a\in (-5;5)$ , which give a total of 11 values of $a$ and ${ b }^{ 2 }\le 25\Leftrightarrow b\in (-5;5)$ , which give a total of 11 values of $b$ . Hence, there is a total of $11\times 11=121$ values of $(a,b)$ pairs which temporarily satisfy the equation.

If we look carefully, for $a=\pm 5$ , $b$ cannot be $\pm 5$ , $\pm 4$ , $\pm 3$ , $\pm 2$ , $\pm 1$ . For $a=\pm 4$ , $b$ cannot be $\pm 5$ , $\pm 4.$ . For $a=\pm 3$ , $a=\pm 2$ and $a=\pm 1$ , $b$ cannot be $\pm 5$ . And finally, for $a=0$ , $b$ can be any integers listed above. So, the number of pairs that temporarily satisfy but not really satisfy the equation is $2\times 10+2\times 4+6\times 2=40$

In conclusion, there are $121-40= \boxed{81}$ pairs of $(a,b)$ which satisfy $\left| a+bi \right| \le 5$

It will come out to a circle of radius 5 units.

Either graph it or use the property that if a point is inside the circle,then a^2+b^2-r^2 is less than or equal to zero.

In each quadrant, you will get 15 points ( excluding the points on co-ordinate axes) .So a total of 60 points will be obtained. A total of 22 points will be on coordinate axes. So total points = 82.

But 0+i0 is not a complex number, hence a total of 81 points will be there.