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JenniferSmart1 2 years ago How do I find absolute values of complex numbers? |7-i|

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1. satellite73

pythatgoras $|a+bi|=\sqrt{a^2+b^2}$

2. satellite73

*pythagoras

3. satellite73

in your case $$a=7,b=-1$$ so you can just about do it in your head

4. satellite73

$\sqrt{7^2+1^2}=\sqrt{491}=\sqrt{50}=5\sqrt2$

5. satellite73

not much to memorize, the hypotenuse is the square root of the sum of the squares, as in a right triangle

6. JenniferSmart1

but what has an "i" to do with a triangle?

7. satellite73

|dw:1358396646228:dw|

8. satellite73

the complex numbers live in the complex plane the absolute value is the distance from the origin, which you get via pythagoras $a^2+b^2=h^2$ $h=\sqrt{a^2+b^2}$

9. JenniferSmart1

is the triangle always in that quadrant?

10. Callisto

No, it depends on the complex number. Let a, b = +ve a+bi => quad. I -a + bi => quad. II -a - bi => quad. III a - bi => quad. IV

11. JenniferSmart1

ahh good to know...so is "a" and "b" the length of the lines?

12. satellite73

it doesn't make any difference however, what quadrant you are in, it is still $|a+bi|=\sqrt{a^2+b^2}$ i just put it there because your number was $$7-i$$

13. satellite73

|dw:1358397280980:dw|

14. Callisto

|dw:1358397320522:dw|

15. JenniferSmart1

That makes sense...I just have one more dumb question a+bi....is that the location of the point at the end of that line?

16. JenniferSmart1

oh no that is the length of the line....durrr..sorry!!!1

17. JenniferSmart1

the absolute value of a+bi is the length of the line...correct?

18. Callisto

a + bi a = number (coordinate) in the real part b = number (coordinate) in the imaginary part | a+bi | = length of the line => yes, I think (Actually... I haven't learnt it in the lesson yet.. So...)

19. JenniferSmart1

that's cool...It makes sense to me though. Thanks soo much @Callisto =)

20. satellite73

yes, it is the length of the line, that is, the distance between the complex number $$a+bi$$ and the origin

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