
#1
Jan1206, 09:56 AM

P: 426

x^2 + 2x + 5 = 0.Find the root of this eqn.can use polar system.
is the answer=(1+2i) @ (12i)?? pls help....... thanx.... 



#2
Jan1206, 10:19 AM

HW Helper
P: 2,566

That answer is in cartesian form. A polar answer looks like r e^{iΘ}. Polar coordinates wouldn't really be appropriate to do this problem, although you could easily put the final answer from cartesian form, z=a+bi, into polar coordinates using r^{2}=a^{2}+b^{2}, Θ=tan^{1}(b/a).




#3
Jan1206, 10:23 AM

P: 426

can youpls show me the stpes??plsssss




#4
Jan1206, 10:34 AM

HW Helper
P: 2,566

polar coordinate
I won't do it for you, but I'll answer any specific questions you have. But if you are going to use cartesion coordinates, just use the quadratic formula.




#5
Jan1206, 10:36 AM

P: 426

i have tried it.so is it the ans is 5^1/2 e(j0.6476pai)??




#6
Jan1206, 02:54 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,904

Please copy the problem exactly as it is given to you. In your first post you said
"x^2 + 2x + 5 = 0.Find the root of this eqn.can use polar system. is the answer=(1+2i) @ (12i)??" "can use polar system" doesn't mean you have to! It's easy to solve the equation by completing the square. Yes, the solutions are x= 1+ 2i and x= 1 2i. Are you required to write the answers in polar form? 



#7
Jan1306, 03:37 PM

P: 17

AFAIK, there's no polar equivalent to adding numbers, so If anyone who knows how to solve it using polar system, please let us know.
Apart from that, it most prolly is to find the roots using quadratic formula in cartesian form ( x+iy ) and convert it to polar form r(e)^iD, where r=sqrt(x^2+y^2) and angle D=arctan(y/x) 



#8
Jan1306, 03:38 PM

P: 426

ya,require.....




#9
Jan1306, 03:48 PM

P: 17

then convert using the expressions I gave in the last post.




#10
Jan1406, 10:23 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,904

Polar form of the number a+ bi is either [itex]r(cos\theta+ i sin\theta)[/itex] or [itex]r e^{i\theta}[/itex] (since [itex]e^{i \theta}= cos\theta+ i sin\theta[/itex] they are equivalent) where r is a+ bi and [itex]\theta[/itex] is the "argument" or angle the line through (0,0) and (a,b) makes with the positive real axis. For a+ bi, [itex]r= \sqrt{a^2+ b^2}[/itex] and [itex]\theta= arctan(\frac{b}{a})[/itex] as long as a is not 0. If a is 0 and b is positive, then [itex]\theta= \frac{\pi}{2}[/itex]. If a is 0 and b is negative, then [itex]\theta= \frac{\pi}{2}[/itex]. The number 0 (0+ 0i) cannot be written in "polar form".
If you were given a problem requiring the answer in polar form, surely you were already taught all of that? 


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