Question

Find four distinct complex numbers (which are neither purely imaginary nor purely real) such that each has an absolute value of 3.

Answer 1

Answer:

• 0.5 + 2.985i
• 1 + 2.828i
• 1.5 + 2.598i
• 2 + 2.236i

Explanation:

Complex numbers have the general form a + bi, where a is the real part and b is the imaginary part.

Since, the numbers are neither purely imaginary nor purely real a ≠ 0 and b ≠ 0.

The absolute value of a complex number is its distance to the origin (0,0), so you use Pythagorean theorem to calculate the absolute value. Calling it |C|, that is:

• $|C| = \sqrt{a^2+b^2}$

Then, the work consists in finding pairs (a,b) for which:

• $\sqrt{a^2+b^2}=3$

You can do it by setting any arbitrary value less than 3 to a or b and solving for the other:

$\sqrt{a^2+b^2}=3\\ \\ a^2+b^2=3^2\\ \\ a^2=9-b^2\\ \\ a=\sqrt{9-b^2}$

I will use b =0.5, b = 1, b = 1.5, b = 2

$b=0.5;a=\sqrt{9-0.5^2}=2.958\\ \\b=1;a=\sqrt{9-1^2}=2.828\\ \\b=1.5;a=\sqrt{9-1.5^2}=2.598\\ \\b=2;a=\sqrt{9-2^2}=2.236$

Then, four distinct complex numbers that have an absolute value of 3 are:

• 0.5 + 2.985i
• 1 + 2.828i
• 1.5 + 2.598i
• 2 + 2.236i