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Easy qn (1 Viewer)

Richard Lee

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There are two questions; find the square roots of 3+4i and find sqrt(3+4i). Do u think the solutions are same?
 
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edit: i should probably leave these questions for the '04ers...
 

sven0023

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hmm ill give it a go

1)

sqr/(3+4i) = x+iy
3+4i = (x+iy)^2
= x^2-y^2+2ixy

Equating real/non real, 3 = x^2-y^2 and y = 2/x

Sub y = 2/x in, 3 = x^2 - (2/x)^2
3x^2 = x^4 - 4
x^4 - 3x^2 - 4 = 0
(x^2-4)(x^2+1) = 0
x^2 = 4
x = +/- 2
y = +/- 1

Therefore, sqr/(3+4i) = 2+i or -2-i


2) is sqr/(3+4i) just the positive values?? ie 2+i
:confused:
 

turtle_2468

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There's no such thing as a "positive" square root in complex numbers... :)
ie the square root symbol in reals implies the positive one, but there's no satisfactory way to imply positiveness in complex, so ppl just write both...
 

Richard Lee

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Originally posted by turtle_2468
There's no such thing as a "positive" square root in complex numbers... :)
ie the square root symbol in reals implies the positive one, but there's no satisfactory way to imply positiveness in complex, so ppl just write both...
Good! So, sqrt(4+0i)=(+/-)(2+0i) and sqrt(4)=2.
I think I know what you mean.
 

KeypadSDM

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Originally posted by Richard Lee
Good! So, sqrt(4+0i)=(+/-)(2+0i) and sqrt(4)=2.
I think I know what you mean.
Actually that's a misinterpretation.

Square roots of complex numbers can't be identified as either positive or negative.

An example: - 3 + 4i OR 3 - 4i

Which is positive, which is negative? Neither has a right to either title, hence the ambiguity.

If the unreal part is zero, then the number is real. sqrt(real) is always positive.
 

spice girl

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in fact i remember chucking a fuss about it in jr high. cos the teacher taught that sqrt(n) meant the number(s) that when squared would give the number n. So to answer sqrt(4) i'd say +/-2. But then *NO*, the teacher says it's only the positive value, and i ask why, and he goes cos there's no negative sign behind it. and i thought "BS! wot an ignorant sweeping assumption"

bleh
 

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