# Thread: Complex number to polarform

1. ## Complex number to polarform

Hello all,

I was hoping for some help on turning complex number to polar form. I'm working on a nonhomogenous difference equation, and while solving I got the roots $\displaystyle \frac{1.5\pm\sqrt{-1.75}}{2}$.

Ok, so far so good. The modulus, $\displaystyle \rho=\sqrt{(\frac{3}{4})^2+(\frac{\sqrt{1,75}}{2}) ^2}$, which gives me $\displaystyle \rho=1$. Then the argument, $\displaystyle \cos \theta = \frac{3/4}{1}=\frac{3}{4}\mbox{, } \sin \theta = \frac{\sqrt{1,75}/2}{1}=\frac{\sqrt{1,75}}{2}$. This gives me $\displaystyle \cos^{-1} \theta =\sin^{-1} \theta = 0.7227342478$. Is this correct?

The root I got, $\displaystyle \frac{1.5\pm\sqrt{-1.75}}{2}$, when I want to find $\displaystyle \sin \theta = \frac{b}{\rho}$, what value of b do I use (because of the $\displaystyle \pm$)? The negative or positive value of $\displaystyle \frac{\sqrt{-1.75}}{2}$?

Is there any way to make the argument a "pretty" $\displaystyle \pi$-fraction (if it exists)?

Thank you for any help!

2. ## complex number

hi, dont think that make any sense something is not right.whats the original question.

3. actually it does make sense sorry i was only glancing afeter i start thinking about it.anyways.let z=3/4 ± i√7/8 (rember that i=√-1 and the √-7/8 is the same as √-1 times √7 ok) polar form =r[cos(θ)+isin(θ)]

[z]=r=(a^2+b^2)=85/64 (note that r is indeed the modulus of z)

arg(z)=tan(θ)=6/7

θ=tan^-1(6/7)
θ=40.60
therefore
arg(z)=40.60 (you can convert to pi radians if you want ok)

now we have everything to represent in polar form so now substitute.

polar form of Z=85/64[cos(40.60) + isin(40.60)]. (the squared root of any -number does not exist thats its imaginary thats why i is used to replace √-n=i ok.) good luck hope this helps and talk to me if you need anything else god be with you.