Question

If z is a complex number such that $\left| z \right|=1$, prove that $\dfrac{z-1}{z+1}$ is purely imaginary.
What will be your conclusion if z = 1?

Answer 1

Hint: Use the fact that if a complex number z is purely imaginary, then $z+\bar{z}=0$. Use the fact that $z\bar{z}={{\left| z \right|}^{2}}$ and $\overline{a+b}=\bar{a}+\bar{b}$ and $\overline{\left( \dfrac{a}{b} \right)}=\dfrac{{\bar{a}}}{{\bar{b}}}$. Consider the expression $\left( \dfrac{z-1}{z+1} \right)+\overline{\left( \dfrac{z-1}{z+1} \right)}$ and use the properties mentioned above to prove that the expression is equal to 0. Hence prove that the complex number $\dfrac{z-1}{z+1}$ is purely imaginary. Infer the result when z =1.

Complete step-by-step answer:
Let $A=\dfrac{z-1}{z+1}$
We have $A+\bar{A}=\left( \dfrac{z-1}{z+1} \right)+\overline{\left( \dfrac{z-1}{z+1} \right)}$
We know that $\overline{\left( \dfrac{a}{b} \right)}=\dfrac{{\bar{a}}}{{\bar{b}}}$
Using the above property of conjugate of complex numbers, we get
$A+\bar{A}=\dfrac{z-1}{z+1}+\dfrac{\overline{z-1}}{\overline{z+1}}$
We know that $\overline{a+b}=\bar{a}+\bar{b}$
Using the above property of conjugate of complex numbers, we get
$A+\bar{A}=\dfrac{z-1}{z+1}+\dfrac{\bar{z}-1}{\bar{z}+1}$
Now, we know that $z\bar{z}={{\left| z \right|}^{2}}=1$
Dividing both sides by z, we get
$\bar{z}=\dfrac{1}{z}$
Hence, we have
$A+\bar{A}=\dfrac{z-1}{z+1}+\dfrac{\dfrac{1}{z}+1}{\dfrac{1}{z}-1}$
Multiplying numerator and denominator of the second fraction by z, we get
$A+\bar{A}=\dfrac{z-1}{z+1}+\dfrac{1+z}{1-z}=\dfrac{z-1}{z+1}-\dfrac{z+1}{z-1}=0$
Hence, we have $A+\bar{A}=0$
Hence, we have A is purely imaginary.
Hence, we have $\dfrac{z-1}{z+1}$ is purely imaginary.
When z = 1, we have $\dfrac{z-1}{z+1}=0$, which is purely imaginary(0 is considered to be both purely imaginary as well as purely real).
Hence the statement holds true for z = 1.

Note: Alternative Solution:
Since $\left| z \right|=1,$ we have $z={{e}^{i\theta }}$
Now, we have $\dfrac{z-1}{z+1}=\dfrac{{{e}^{i\theta }}-1}{{{e}^{i\theta }}+1}$
Multiplying the numerator and denominator of RHS by ${{e}^{-\dfrac{i\theta }{2}}}$, we get
$\dfrac{z-1}{z+1}=\dfrac{{{e}^{\dfrac{i\theta }{2}}}-{{e}^{-\dfrac{i\theta }{2}}}}{{{e}^{\dfrac{i\theta }{2}}}+{{e}^{-\dfrac{i\theta }{2}}}}$
We know that ${{e}^{ix}}+{{e}^{-ix}}=2\cos x$ and ${{e}^{ix}}-{{e}^{-ix}}=2i\sin x$
Hence, we have
$\dfrac{z-1}{z+1}=\dfrac{2i\sin \left( \dfrac{\theta }{2} \right)}{2\cos \left( \dfrac{\theta }{2} \right)}=i\tan \left( \dfrac{\theta }{2} \right)$, which is purely imaginary.
Hence, we have
$\dfrac{z-1}{z+1}$ is purely imaginary.