Q:

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

Accepted Solution

A:
Answer:0.5 + 2.985i1 + 2.828i1.5 + 2.598i2 + 2.236iExplanation: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 β‰  0 and b β‰  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:[tex]|C| = \sqrt{a^2+b^2}[/tex]Then, the work consists in finding pairs (a,b) for which:[tex]\sqrt{a^2+b^2}=3[/tex]You can do it by setting any arbitrary value less than 3 to a or b and solving for the other:[tex]\sqrt{a^2+b^2}=3\\ \\ a^2+b^2=3^2\\ \\ a^2=9-b^2\\ \\ a=\sqrt{9-b^2}[/tex]I will use b =0.5, b = 1, b = 1.5, b = 2[tex]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[/tex]Then, four distinct complex numbers that have an absolute value of 3 are:0.5 + 2.985i1 + 2.828i1.5 + 2.598i2 + 2.236i