Eleven Standard >> Complex number | An important illustration

Important complex number problem

llustraion 1. Express \(\frac{1}{1-\cos\theta+2i\sin\theta}\) in the form A+iB. Hence sho that A=\(\frac{1}{5+3\cos\theta}\) and B=\(\frac{-2\cot\frac{\theta}{2}}{5+3\cos\theta}\)

Solution:
z=\(\frac{1}{(1-\cos\theta)+2i\sin\theta}\)
  =\(\frac{(1-\cos\theta)-2i\sin\theta}{(1-\cos\theta)^{2}+(2i\sin\theta)^{2}}\)
  =\(\frac{1-\cos\theta-2i\sin\theta}{(2\sin^{2}\frac{\theta}{2})^{2}-4(-1)\sin^{2}\theta}\)
  =\(\frac{1-\cos\theta-2i\sin\theta}{(2\sin^{2}\frac{\theta}{2})^{2}+4(2\sin\frac{\theta}{2}\cos\frac{\theta}{2})^{2}}\)
  =\(\frac{2\sin^{2}\frac{\theta}{2}-2i.2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^{2}\frac{\theta}{2}[2\sin^{2}\frac{\theta}{2}+8\cos^{2}\frac{\theta}{2}]}\)
   =\(\frac{2\sin^{2}\frac{\theta}{2}[1-2i\cot\frac{\theta}{2}]}{2\sin^{2}\frac{\theta}{2}[2+6\cos^{2}\frac{\theta}{2}]}\)
   =\(\frac{1-2i\cot\frac{\theta}{2}}{2+3(1+\cos\theta)}\)
   =\(\frac{1}{5+3\cos\theta}+i.\frac{-2\cot\frac{\theta}{2}}{5+3\cos\theta}\)
   =A+iB
Hence, A=\(\frac{1}{5+3\cos\theta}\), B=\(\frac{-2\cot\frac{\theta}{2}}{5+3\cos\theta}\)

Illustration 2. Find real \(\theta\) such that \(\frac{3+2i\sin\theta}{1-2i\sin\theta}\)

Solution:
Let z=\(\frac{3+2i\sin\theta}{1-2i\sin\theta}\)
      =\(\frac{(3+2i\sin\theta)(1+2i\sin\theta)}{(1-2i\sin\theta)(1+2i\sin\theta)}\)
      =\(\frac{3(1+2i\sin\theta)+2i\sin\theta(1+2i\sin\theta)}{(1+4\sin^{2}\theta)}\)
      =\(\frac{3+6i\sin\theta+2i\sin\theta-4\sin^{2}\theta)}{(1+4\sin^{2}\theta)}\)
       =\(\frac{3-4\sin^{2}\theta}{1+4\sin^{2}\theta}\)+i\(\frac{8\sin\theta}{1+4\sin^{2}\theta}\)
       
For z purely real, \(I_{M}(z)\)=0
i.e, \(\frac{8\sin\theta}{1+4\sin^{2}\theta}\)=0
\(\Rightarrow\) \(8\sin\theta\)=0
\(\Rightarrow\) \(\sin\theta\)=0
\(\Rightarrow\) \(\sin\theta\)=0
\(\Rightarrow\) \(\theta\)=\(n\pi\) [where \(n\in I\)]