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

4 0

4 years ago

A password is 4 characters long, consisting of 2 letters and 2 numbers. The password must begin and end with a

slamgirl [31]

Answer:

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Which graph solves the following system? x+2y=4 5x−2y=8

12345 [234]

Answer:

elimination method

x+2y=4 1

5x-2y=8 2

1+2

6x=12

x=2

plug into x+2y=4

2+2y=4

2y=4-2

2y=2

y=1

(2,1)

so graph 1

6 0

3 years ago