Question

8. If tan (a + iB) = x + iy, then prove that
(a) x² + y2 + 2x cot 2a =1 (b)​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tan(A+i\,B)=x+i\,y$

$\implies\,\dfrac{1}{i}\cdot\dfrac{{e}^{i\,A-B}-{e}^{-i\,A+B}}{{e}^{i\,A-B}+{e}^{-i\,A+B}}=x+i\,y$

$\implies\,\dfrac{{e}^{i\,A-B}-{e}^{-i\,A+B}}{{e}^{i\,A-B}+{e}^{-i\,A+B}}=i\,x-y$

$\implies\,\dfrac{{e}^{-B}\cdot{e}^{i\,A}-{e}^{B}\cdot{e}^{-i\,A}}{{e}^{-B}\cdot{e}^{i\,A}+{e}^{B}\cdot{e}^{-i\,A}}=i\,x-y$

Using componendo and dividendo,

$\implies\,\dfrac{{e}^{-B}\cdot{e}^{i\,A}-{e}^{B}\cdot{e}^{-i\,A}+{e}^{-B}\cdot{e}^{i\,A}+{e}^{B}\cdot{e}^{-i\,A}}{{e}^{-B}\cdot{e}^{i\,A}-{e}^{B}\cdot{e}^{-i\,A}-{e}^{-B}\cdot{e}^{i\,A}-{e}^{B}\cdot{e}^{-i\,A}}=\dfrac{i\,x-y+1}{i\,x-y-1}$

$\implies\,\dfrac{{e}^{-B}\cdot{e}^{i\,A}+{e}^{-B}\cdot{e}^{i\,A}}{-{e}^{B}\cdot{e}^{-i\,A}-{e}^{B}\cdot{e}^{-i\,A}}=\dfrac{i\,x-y+1}{i\,x-y-1}$

$\implies\,-\dfrac{{e}^{-B}\cdot{e}^{i\,A}}{{e}^{B}\cdot{e}^{-i\,A}}=\dfrac{-(y-1)+i\,x}{-(y+1)+i\,x}$

$\implies\,-\dfrac{{e}^{i(2A)}}{{e}^{2B}}=\dfrac{(y-1)-i\,x}{(y+1)-i\,x}$

$\implies\,-\dfrac{\cos(2A)+i\sin(2A)}{{e}^{2B}}=\dfrac{\big\{(y-1)-i\,x\big\}\big\{(y+1)+i\,x\big\}}{\big\{(y+1)-i\,x\big\}\big\{(y+1)+i\,x\big\}}$

$\implies\,-\dfrac{\cos(2A)+i\sin(2A)}{{e}^{2B}}=\dfrac{(y-1)(y+1)-i\,x(y+1)-i\,x(y-1)-{i}^{2}\,{x}^{2}}{\left(y+1\right)^{2}+{x}^{2}}$

$\implies\,-\dfrac{\cos(2A)}{{e}^{2B}}-i\,\dfrac{\sin(2A)}{{e}^{2B}}=\dfrac{{y}^{2}-1-i\,x(y+1+y-1)+{x}^{2}}{\left(y+1\right)^{2}+{x}^{2}}$

$\implies\,-\dfrac{\cos(2A)}{{e}^{2B}}-i\,\dfrac{\sin(2A)}{{e}^{2B}}=\dfrac{{y}^{2}+{x}^{2}-1-i\cdot2xy}{\left(y+1\right)^{2}+{x}^{2}}$

$\implies\,-\dfrac{\cos(2A)}{{e}^{2B}}-i\,\dfrac{\sin(2A)}{{e}^{2B}}=-\dfrac{1-{y}^{2}-{x}^{2}}{\left(y+1\right)^{2}+{x}^{2}}-i\,\dfrac{2xy}{\left(y+1\right)^{2}+{x}^{2}}$

On comparing the real and imaginary parts,

$\implies\,\dfrac{\cos(2A)}{{e}^{2B}}=\dfrac{1-{y}^{2}-{x}^{2}}{\left(y+1\right)^{2}+{x}^{2}}\,\,\,\,\,\,\,\,\,\&\,\,\,\,\,\,\,\,\dfrac{\sin(2A)}{{e}^{2B}}=\dfrac{2xy}{\left(y+1\right)^{2}+{x}^{2}}$

Dividing both equations,

$\implies\,\dfrac{\dfrac{\cos(2A)}{{e}^{2B}}}{\dfrac{\sin(2A)}{{e}^{2B}}}=\dfrac{\dfrac{1-{y}^{2}-{x}^{2}}{\left(y+1\right)^{2}+{x}^{2}}}{\dfrac{2xy}{\left(y+1\right)^{2}+{x}^{2}}}$

$\implies\,\dfrac{\cos(2A)}{\sin(2A)}=\dfrac{1-{y}^{2}-{x}^{2}}{2xy}$

$\implies\,\cot(2A)=\dfrac{1-{y}^{2}-{x}^{2}}{2xy}$

$\implies\,2xy\cdot\cot(2A)=1-{y}^{2}-{x}^{2}$

$\implies\,{x}^{2}+{y}^{2}+2xy\cdot\cot(2A)=1$