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Find all complex numbers z satisfying the equation (z+1)/(z-1) = i

waffles  Oct 29, 2017

Best Answer 

 #1
avatar+20680 
+1

Find all complex numbers z satisfying the equation (z+1)/(z-1) = i

 

\(\begin{array}{|lrcll|} \hline & \dfrac{z+1}{z-1} &=& i \quad & | \quad z = a+bi \\\\ & \dfrac{a+bi+1}{a+bi-1} &=& i \\\\ & a+bi+1 &=& i(a+bi-1) \\\\ & a+bi+1 - i(a+bi-1)&=& 0 \\\\ & a+bi+1 - ia-bi^2+i&=& 0 \quad & | \quad i^2 = -1\\ \\ & a+bi+1 - ia+b+i&=& 0 \\ \\ & a+bi+1 - ia+b+i&=& 0 \\ \\ & ( a+b+1) + (b-a+1) i &=& 0 \\ \\ \hline (1) & (a+b+1) &=& 0 \\ (2) & (b-a+1) &=& 0 \\ \hline (1) + (2) & 2b+2 &=& 0 \quad & | \quad : 2 \\ & b+1 &=& 0 \\ & \mathbf{b} &\mathbf{=}& \mathbf{-1} \\ \hline (1) - (2) & 2a &=& 0 \quad & | \quad : 2 \\ & \mathbf{a} &\mathbf{=}& \mathbf{0} \\ \hline \end{array}\)


\(z = a+bi \\ z = 0 -i \\ \mathbf{z = -i}\)

 

laugh

heureka  Oct 30, 2017
 #1
avatar+20680 
+1
Best Answer

Find all complex numbers z satisfying the equation (z+1)/(z-1) = i

 

\(\begin{array}{|lrcll|} \hline & \dfrac{z+1}{z-1} &=& i \quad & | \quad z = a+bi \\\\ & \dfrac{a+bi+1}{a+bi-1} &=& i \\\\ & a+bi+1 &=& i(a+bi-1) \\\\ & a+bi+1 - i(a+bi-1)&=& 0 \\\\ & a+bi+1 - ia-bi^2+i&=& 0 \quad & | \quad i^2 = -1\\ \\ & a+bi+1 - ia+b+i&=& 0 \\ \\ & a+bi+1 - ia+b+i&=& 0 \\ \\ & ( a+b+1) + (b-a+1) i &=& 0 \\ \\ \hline (1) & (a+b+1) &=& 0 \\ (2) & (b-a+1) &=& 0 \\ \hline (1) + (2) & 2b+2 &=& 0 \quad & | \quad : 2 \\ & b+1 &=& 0 \\ & \mathbf{b} &\mathbf{=}& \mathbf{-1} \\ \hline (1) - (2) & 2a &=& 0 \quad & | \quad : 2 \\ & \mathbf{a} &\mathbf{=}& \mathbf{0} \\ \hline \end{array}\)


\(z = a+bi \\ z = 0 -i \\ \mathbf{z = -i}\)

 

laugh

heureka  Oct 30, 2017

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