SOLUTION: Hello, I am having trouble with absolute value of complex numbers. I have 5 complex numbers and need to figure out which one has the greatest absolute value. 1)-2+i sqrt(10) 2)3-

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: Hello, I am having trouble with absolute value of complex numbers. I have 5 complex numbers and need to figure out which one has the greatest absolute value. 1)-2+i sqrt(10) 2)3-       Log On


   

Question 1021320: Hello, I am having trouble with absolute value of complex numbers. I have 5 complex numbers and need to figure out which one has the greatest absolute value. 1)-2+i sqrt(10) 2)3- i sqrt (5) 3) - sqrt(9)-2i (I think this answer is sqrt (13) 4) 2+3i (I think this answer is also sqrt (13) and 5)sqrt (6)-3i I appreciate any help anyone can give me!
Found 2 solutions by Fombitz, Theo:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
For a complex number, Z=a%2Bbi, the magnitude is,
abs%28Z%29=sqrt%28a%5E2%2Bb%5E2%29
.
.
(-2,sqrt%2810%29):abs%28Z%29=sqrt%28%28-2%29%5E2%2B%28sqrt%2810%29%29%5E2%29=sqrt%284%2B10%29=sqrt%2814%29
(3,-sqrt%285%29):abs%28Z%29=sqrt%28%283%29%5E2%2B%28-sqrt%285%29%29%5E2%29=sqrt%289%2B5%29=sqrt%2814%29
(-sqrt%289%29,-2):abs%28Z%29=sqrt%28%28-sqrt%289%29%29%5E2%2B%28-2%29%5E2%29=sqrt%289%2B4%29=sqrt%2813%29
(2,3):abs%28Z%29=sqrt%28%282%29%5E2%2B%283%29%5E2%29=sqrt%284%2B9%29=sqrt%2813%29
(sqrt%286%29,-3):abs%28Z%29=sqrt%28%28sqrt%286%29%29%5E2%2B%28-3%29%5E2%29=sqrt%286%2B9%29=sqrt%2815%29

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
i believe this is what you are looking for.

http://www.regentsprep.org/regents/math/algtrig/ato6/absvlecomlesson.htm

let's do them in turn.

remember the general form is a + bi.
this can also be shown as a + ib, as in some of your problems.
a is the real part.
b is the coefficient of the imaginary part.

the absolute value is equal to sqrt(a^2 + b^2).

your first one is:

1) -2 + i * sqrt(10)

a = -2
b = sqrt(10)

sqrt(a^2 + b^2) = sqrt(4 + 10) = sqrt(14)

-----

2) 3 - i * sqrt (5)

a = 3
b = -sqrt(5)

sqrt(a^2 + b^2) = sqrt(9 + 5) = sqrt(14)

-----

3) - sqrt(9) - 2i

a = sqrt(9)
b = -2

sqrt(a^2 + b^2) = sqrt(9 + 4) = sqrt(13)

-----

4) 2 + 3i

a = 2
b = 3

sqrt(a^2 + b^2) = sqrt(4 + 9) = sqrt(13)

-----

5) sqrt(6)-3i

a = sqrt(6)
b = -3

sqrt(a^2 + b^2) = sqrt(6 + 9) = sqrt(15)

-----

your solutions are:

1. sqrt(14)
2. sqrt(14)
3. sqrt(13)
4. sqrt(13)
5. sqrt(15)

looks like the last one is the biggest.

looks like you did some of the calculations correctly.