Given complex number is z = \(\frac{1+2i}{1-3i}\)
We know that the polar form of a complex number Z=x+iy is given by Z=|Z|(cosθ+isinθ)
Where,
|Z|=modulus of complex number= \(\sqrt{x^2+y^2}\)
θ =arg(z)=argument of complex number= tan-1 \(\Big(\frac{|y|}{|x|}\Big)\)
Now for the given problem,
⇒ |z| = \(\sqrt{\Big(-\frac{1}{2}\Big)^2+\Big(\frac{1}{2}\Big)^2}\)
⇒ |z| = \(\sqrt{\frac{1}{4}+\frac{1}{4}}\)
⇒ |z| = \(\sqrt{\frac{1}{2}}\)
⇒ |z| = \(\frac{1}{\sqrt{2}}\)
Since x>0,y>0 complex number lies in 2nd quadrant and the value of θ will be as follows 90°≤θ≤180°.
⇒ θ = tan-1(1)
⇒ θ = \(\frac{3\pi}{4}\)
⇒ z = \(\frac{1}{\sqrt{2}}\Big(cos\Big(\frac{3\pi}{4}\Big) + isin\Big(\frac{3\pi}{4}\Big)\Big)\)
∴ The Polar form of Z=\(\frac{1+2i}{1-2i}\) is z =\(\frac{1}{\sqrt{2}}\Big(cos\Big(\frac{3\pi}{4}\Big) - isin\Big(\frac{3\pi}{4}\Big)\Big)\)