Question

If ( z - 1 by z ) = 6, then ffind the value of ( z + 1 by z )​

Answer 1

Given :-

${\sf{\ \ z - {\dfrac{1}{z}} = 6}}$

To Find :-

${\sf{\ \ z + {\dfrac{1}{z}} }}$

Solution :-

Squaring both sides of ${\sf{z - {\dfrac{1}{z}} . }}$

$\implies{\sf{ ( z - {\dfrac{1}{z}} ) ^2 = (6)^2 }}$

${\boxed{\tt{\bigstar \ \ Identity \ : \ (a - b)^2 = a^2 + b^2 - 2ab }}}$

${\tt{Here, \ a = z, \ b = {\dfrac{1}{z}}}}$

$\implies{\sf{ (z)^2 + ( {\dfrac{1}{z}} ) ^2 - 2 . z . {\dfrac{1}{z}} = 36}}$

$\implies{\sf{ z^2 + {\dfrac{1}{z^2}} - 2 = 36}}$

$\implies{\sf{ z^2 + {\dfrac{1}{z^2}} = 36 + 2}}$

$\implies{\sf{ z^2 + {\dfrac{1}{z^2}} = 38 \qquad \ \ ...(1) }}$

Now,

Squaring of ${\sf{z + {\dfrac{1}{z}} . }}$

$\implies{\sf{ ( z + {\dfrac{1}{z}} )^2 }}$

${\boxed{\tt{\bigstar \ \ Identity \ : \ (a + b)^2 = a^2 + b^2 + 2ab}}}$

${\tt{Here, \ a = z, \ b = {\dfrac{1}{z}}}}$

$\implies{\sf{ (z)^2 + ( {\dfrac{1}{z}} )^2 + 2 . z . {\dfrac{1}{z}} }}$

$\implies{\sf{z^2 + {\dfrac{1}{z^2}} + 2 }}$

$\implies{\sf{38 + 2 \qquad \ \ ...from (1) }}$

$\implies{\boxed{\boxed{\sf{40}}}}$

Now,

We get : ${\sf{ ( z + {\dfrac{1}{z}} ) ^2 = 40}}$

$\implies{\sf{ z + {\dfrac{1}{z}} = {\sqrt{40}}}}$

Answer 2

$\tt{Squaring\: both \: sides \: of \:z-\frac{1}{z}}$

$\implies$$\tt{{(z - \frac{1}{z} )}^{2} = {(6)}^{2} }$

$\tt\red{Identity={(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab}$

$\tt{Here,a=z,b=\frac{1}{z}}$

$\implies$$\tt{{(z)}^{2} + {( \frac{1}{z}) }^{2} - 2.z. \frac{1}{z} = 36}$

$\implies$$\tt{ {z}^{2} + \frac{1}{ {z}^{2} } - 2 = 36}$

$\implies$$\tt{ {z}^{2} + \frac{1}{ {z}^{2} } = 36 + 2}$

$\implies$$\tt{ {z}^{2} + \frac{1}{ {z}^{2} } = 38.......(1)}$

$\tt{Now,squaring \: of \: z+\frac{1}{z}}$

$\implies$$\tt{{ {(z} + \frac{1}{z} )^{2}}}$

$\tt\red{Identity={(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab}$

$\implies$$\tt{Here ,\: a = z \:, b = \frac{1}{z} }$

$\implies$$\tt{ {(z)}^{2} + \ {( \frac{1}{z}) }^{2} + 2.z. \frac{1}{z} }$

$\implies$$\tt{ {z}^{2} + \frac{1}{ {z}^{2} } + 2}$

$\implies$$\tt{38+2......from(1)}$

$\implies$$\tt{40}$

$\tt{Now,}$

$\tt{we\: get:{(z + \frac{1}{z} )}^{2} = 40}$

$\implies$$\tt\red{z + \frac{1}{z} = \sqrt{40} }$