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# Really need help on this problem

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Prove that if w,z are complex numbers such that |w|=|z|=1 and wz≠1, then {w+z}/{1+wz} is a real number. How would I prove this using the complex plane?

Guest Mar 1, 2018
#1
+20016
+1

I assume:

Prove that if $$w,z$$ are complex numbers such that $$|w|=|z|=1$$

and  $$wz\ne -1$$, then

$$\frac{w+z}{1+wz}$$ is a real number.

$$\text{Let w = \cos(\theta)+ \imath \sin(\theta) } \\ \text{Let z = \cos(\phi) + \imath\sin(\phi) }$$

$$\begin{array}{|rcll|} \hline w+z &=& \cos(\theta)+ \imath \sin(\theta) + \cos(\phi) + \imath\sin(\phi) \\ &=& \Big(\cos(\theta)+ \cos(\phi)\Big) + \imath \Big( \sin(\theta) + \sin(\phi) \Big) \\\\ w\cdot z &=& \Big(\cos(\theta)+ \imath \sin(\theta) \Big) \Big( \cos(\phi) + \imath\sin(\phi) \Big) \\ &=& \cos(\theta)\cos(\phi) + \imath\sin(\phi)\cos(\theta) + \imath \sin(\theta)\cos(\phi) + \imath^2 \sin(\theta)\sin(\phi) \\ &=& \cos(\theta)\cos(\phi)+ \imath^2 \sin(\theta)\sin(\phi) + \imath \Big(\sin(\theta)\cos(\phi) + \sin(\phi)\cos(\theta) \Big) \\ && | \quad \imath^2 = -1 \\ &=& \underbrace{\cos(\theta)\cos(\phi)- \sin(\theta)\sin(\phi)}_{=\cos(\theta+\phi)} + \imath \Big(\underbrace{\sin(\theta)\cos(\phi) + \sin(\phi)\cos(\theta)}_{=\sin(\theta+\phi)} \Big) \\ &=& \cos(\theta+\phi) + \imath \sin(\theta+\phi) \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline && \dfrac{w+z}{1+wz} \\\\ &=& \dfrac{ \Big(\cos(\theta)+ \cos(\phi)\Big) + \imath \Big( \sin(\theta) + \sin(\phi) \Big) } { 1+\cos(\theta+\phi) + \imath \sin(\theta+\phi) } \\\\ &=& \dfrac{ \Big(\cos(\theta)+ \cos(\phi)\Big) + \imath \Big( \sin(\theta) + \sin(\phi) \Big) } { \Big(1+\cos(\theta+\phi)\Big) + \imath \sin(\theta+\phi) } \\ && | \quad \small{\text{ Multiply top and bottom by the conjugate}} \\\\ &=& \left[ \dfrac{\Big(\cos(\theta)+ \cos(\phi)\Big) + \imath \Big( \sin(\theta) + \sin(\phi) \Big) } { \Big(1+\cos(\theta+\phi)\Big) + \imath \sin(\theta+\phi) } \right] \left[ \dfrac{\Big(1+\cos(\theta+\phi)\Big) - \imath \sin(\theta+\phi) } { \Big(1+\cos(\theta+\phi)\Big) - \imath \sin(\theta+\phi) } \right] \\\\ &=& \dfrac{ \left[\Big(\cos(\theta)+ \cos(\phi)\Big) + \imath \Big( \sin(\theta) + \sin(\phi) \Big)\right] \left[\Big(1+\cos(\theta+\phi)\Big) - \imath \sin(\theta+\phi) \right] } { \Big(1+\cos(\theta+\phi)\Big)^2 - \imath^2 \sin^2(\theta+\phi) } \\\\ &=& \dfrac{ \left[\Big(\cos(\theta)+ \cos(\phi)\Big) + \imath \Big( \sin(\theta) + \sin(\phi) \Big)\right] \left[\Big(1+\cos(\theta+\phi)\Big) - \imath \sin(\theta+\phi) \right] } { \Big(1+\cos(\theta+\phi)\Big)^2 + \sin^2(\theta+\phi) } \\\\ &=& \dfrac{\Big(\cos(\theta)+ \cos(\phi)\Big)\Big(1+\cos(\theta+\phi)\Big) -\Big(\cos(\theta)+ \cos(\phi)\Big) \imath \sin(\theta+\phi) \\ +\imath \Big( \sin(\theta) + \sin(\phi) \Big)\Big(1+\cos(\theta+\phi)\Big) -\imath \Big( \sin(\theta) + \sin(\phi) \Big) \imath \sin(\theta+\phi) } { \Big(1+\cos(\theta+\phi)\Big)^2 + \sin^2(\theta+\phi) } \\\\ &=& \dfrac{\Big(\cos(\theta)+ \cos(\phi)\Big)\Big(1+\cos(\theta+\phi)\Big) -\imath \Big(\cos(\theta)+ \cos(\phi)\Big) \sin(\theta+\phi) \\ +\imath \Big( \sin(\theta) + \sin(\phi) \Big)\Big(1+\cos(\theta+\phi)\Big) -\imath^2 \Big( \sin(\theta) + \sin(\phi) \Big)\sin(\theta+\phi) } { \Big(1+\cos(\theta+\phi)\Big)^2 + \sin^2(\theta+\phi) } \\\\ &=& \dfrac{\Big(\cos(\theta)+ \cos(\phi)\Big)\Big(1+\cos(\theta+\phi)\Big) + \Big( \sin(\theta) + \sin(\phi) \Big)\sin(\theta+\phi) \\ -\imath \Big(\cos(\theta)+ \cos(\phi)\Big) \sin(\theta+\phi) +\imath \Big( \sin(\theta) + \sin(\phi) \Big)\Big(1+\cos(\theta+\phi)\Big) } { \Big(1+\cos(\theta+\phi)\Big)^2 + \sin^2(\theta+\phi) } \\\\ &=& \dfrac{\Big(\cos(\theta)+ \cos(\phi)\Big)\Big(1+\cos(\theta+\phi)\Big) + \Big( \sin(\theta) + \sin(\phi) \Big)\sin(\theta+\phi) \\ +\imath \left[ \Big( \sin(\theta) + \sin(\phi) \Big)\Big(1+\cos(\theta+\phi)\Big) - \Big(\cos(\theta)+ \cos(\phi)\Big) \sin(\theta+\phi) \right] } { \Big(1+\cos(\theta+\phi)\Big)^2 + \sin^2(\theta+\phi) } \\ \hline \end{array}$$

The imaginary part of the top line, showing that it's equal to zero:

$$\begin{array}{|rcll|} \hline \imath \left[ \underbrace{\Big( \sin(\theta) + \sin(\phi) \Big)\Big(1+\cos(\theta+\phi)\Big) - \Big(\cos(\theta)+ \cos(\phi)\Big) \sin(\theta+\phi) }_{= 0\ ?}\right] \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline && \Big( \sin(\theta) + \sin(\phi) \Big)\Big(1+\cos(\theta+\phi)\Big) -\Big(\cos(\theta)+ \cos(\phi)\Big) \sin(\theta+\phi) \\\\ &=& \sin(\theta) + \sin(\phi) \\ &\qquad +& \sin(\theta)\cos(\theta+\phi) \\ &\qquad +& \sin(\phi)\cos(\theta+\phi) \\ &\qquad -& \cos(\theta)\sin(\theta+\phi) \\ &\qquad -& \cos(\phi)\sin(\theta+\phi) \\\\ &\qquad & \boxed{\cos(\theta+\phi) = \cos(\theta)\cos(\phi)- \sin(\theta)\sin(\phi) \\ \sin(\theta+\phi) = \sin(\theta)\cos(\phi) + \sin(\phi)\cos(\theta)} \\\\ &=& \sin(\theta) + \sin(\phi) \\ &\qquad +& \sin(\theta)\Big( \cos(\theta)\cos(\phi)- \sin(\theta)\sin(\phi) \Big) \\ &\qquad +& \sin(\phi)\Big( \cos(\theta)\cos(\phi)- \sin(\theta)\sin(\phi) \Big) \\ &\qquad -& \cos(\theta)\Big( \sin(\theta)\cos(\phi) + \sin(\phi)\cos(\theta) \Big) \\ &\qquad -& \cos(\phi)\Big( \sin(\theta)\cos(\phi) + \sin(\phi)\cos(\theta) \Big) \\\\ &=& \sin(\theta) + \sin(\phi) \\ &\qquad +& \sin(\theta)\cos(\theta)\cos(\phi)- \sin(\theta)\sin(\theta)\sin(\phi) \\ &\qquad +& \sin(\phi)\cos(\theta)\cos(\phi)- \sin(\phi)\sin(\theta)\sin(\phi) \\ &\qquad -& \cos(\theta)\sin(\theta)\cos(\phi) - \cos(\theta)\sin(\phi)\cos(\theta) \\ &\qquad -& \cos(\phi)\sin(\theta)\cos(\phi) - \cos(\phi)\sin(\phi)\cos(\theta) \\\\ &=& \sin(\theta) + \sin(\phi) \\ &\qquad +& \sin(\theta)\cos(\theta)\cos(\phi)- \sin^2(\theta)\sin(\phi) \\ &\qquad +& \sin(\phi)\cos(\theta)\cos(\phi)- \sin^2(\phi)\sin(\theta) \\ &\qquad -& \cos(\theta)\sin(\theta)\cos(\phi) - \cos^2(\theta)\sin(\phi) \\ &\qquad -& \cos(\phi)\sin(\phi)\cos(\theta) - \cos^2(\phi)\sin(\theta) \\\\ &=& \sin(\theta) + \sin(\phi) \\ &\qquad +& -\sin(\theta)\Big( \underbrace{\sin^2(\phi)+ \cos^2(\phi)}_{=1} \Big) \\ &\qquad +& - \sin(\phi)\Big( \underbrace{\sin^2(\theta)+\cos^2(\theta)}_{=1} \Big) \\ &\qquad +& \underbrace{\sin(\theta)\cos(\theta)\cos(\phi) - \sin(\theta)\cos(\theta)\cos(\phi)}_{=0} \\ &\qquad +& \underbrace{\sin(\phi)\cos(\phi)\cos(\theta) - \sin(\phi)\cos(\phi)\cos(\theta)}_{=0} \\\\ &=& \sin(\theta) + \sin(\phi) -\sin(\theta) - \sin(\phi) \\ &=& 0 \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \imath \left[ Big( \sin(\theta) + \sin(\phi) \Big)\Big(1+\cos(\theta+\phi)\Big) - \Big(\cos(\theta)+ \cos(\phi)\Big) \sin(\theta+\phi) \right] = 0 \\ \hline \end{array}$$

heureka  Mar 2, 2018
edited by heureka  Mar 2, 2018
#2
+20016
0

2. Solution:

I asume:
Prove that if $$w,z$$ are complex numbers such that $$|w|=|z|=1$$
and
$$wz\ne -1$$, then
$$\frac{w+z}{1+wz}$$ is a real number.

$$\text{Let w = a+\imath b \qquad a^2+b^2 = 1\quad (|w|=1) } \\ \text{Let z = c+\imath d \qquad c^2+d^2 = 1\quad (|z|=1)  }$$

$$\begin{array}{|rcll|} \hline w+z &=& a+\imath b + c+\imath d \\ &=& (a+c) + \imath( b + d) \\\\ w\cdot z &=& (a+\imath b) + (c+\imath d) \\ &=& (ac-bd)+\imath(ad+bc) \\ \hline \end{array}$$

$$\small{ \begin{array}{|rcll|} \hline && \dfrac{w+z}{1+wz} \\\\ &=& \dfrac{ (a+c) + \imath( b + d) } { 1+(ac-bd)+\imath(ad+bc) } \\\\ &=& \dfrac{ (a+c) + \imath( b + d) } { (1+ac-bd)+\imath(ad+bc) } \\ && | \quad \small{\text{ Multiply top and bottom by the conjugate}} \\\\ &=& \left[ \dfrac{ (a+c) + \imath( b + d) } { (1+ac-bd)+\imath(ad+bc) } \right] \left[ \dfrac{(1+ac-bd)-\imath(ad+bc)} { (1+ac-bd)-\imath(ad+bc) } \right] \\\\ &=& \dfrac{ \left[(a+c) + \imath( b + d)\right] \left[(1+ac-bd)-\imath(ad+bc) \right] } { (1+ac-bd)^2 - \imath^2 (ad+bc)^2 } \\\\ && \quad | \quad \imath^2=-1 \\\\ &=& \dfrac{ (a+c)(1+ac-bd) -\imath(a+c)(ad+bc) +\imath( b + d)(1+ac-bd)-\imath( b + d)\imath(ad+bc) } { (1+ac-bd)^2 + (ad+bc)^2 } \\\\ &=& \dfrac{ (a+c)(1+ac-bd) -\imath(a+c)(ad+bc) +\imath2( b + d)(1+ac-bd)-\imath^2( b + d)(ad+bc) } { (1+ac-bd)^2 + (ad+bc)^2 } \\\\ && \quad | \quad \imath^2=-1 \\\\ &=& \dfrac{ (a+c)(1+ac-bd) -\imath(a+c)(ad+bc) +\imath2( b + d)(1+ac-bd)+( b + d)(ad+bc) } { (1+ac-bd)^2 + (ad+bc)^2 } \\\\ &=& \dfrac{ (a+c)(1+ac-bd)+( b + d)(ad+bc) + \imath \Big(( b + d)(1+ac-bd)-(a+c)(ad+bc) \Big) } { (1+ac-bd)^2 + (ad+bc)^2 } \\ \hline \end{array} }$$

The imaginary part of the top line, showing that it's equal to zero:

$$\begin{array}{|rcll|} \hline \imath \left[ \underbrace{ ( b + d)(1+ac-bd)-(a+c)(ad+bc) }_{= 0\ ?}\right] \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline && (b + d)(1+ac-bd)-(a+c)(ad+bc) \\\\ &=& b+abc-b^2d+d+acd-d^2b -(a^2d+abc+acd+c^2b ) \\ &=& b+\not{ab}c-b^2d+d+\not{acd}-d^2b -a^2d-\not{abc}-\not{acd}-c^2b \\ &=& b-b^2d+d -d^2b -a^2d -c^2b \\ &=& b+d -b(\underbrace{c^2+d^2}_{=1})-d(\underbrace{a^2+b^2}_{=1}) \\ &=& b+d -b -d \\ &=& 0 \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \imath \left[ ( b + d)(1+ac-bd)-(a+c)(ad+bc) \right] = 0 \\ \hline \end{array}$$

heureka  Mar 2, 2018
#3
0

Could you explain the last 3 steps in the second part of #2 solution? Im not sure how you got the two.

Guest Mar 2, 2018
edited by Guest  Mar 2, 2018
edited by Guest  Mar 2, 2018