Question

If (z-1/z)=6 find the value of :
(z+1/z)​

Answer 1

Answer:

${\sf{ {\sqrt{40}}}}$

Step-by-step explanation:

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

To Find : ${\sf{\ \ z + {\dfrac{1}{z}} }}$

Solution :

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

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

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

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

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

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

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

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

Now,

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

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

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

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

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

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

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

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

Now,

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

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

Answer 2

❏ Question

If (z-1/z)=6 find the value of (z+1/z) .

❏Solution

Given:-

• ( z - 1/z ) = 6 .......(1)

Find:-

• (z + 1/z ) = ?

Explanation:-

we Know,

★ ( a - b)² = (a² + b² -2ab )

So,

So,➠ ( z - 1/z )² = (z² + 1/z² - 2 )

( keep Value by (1) )

➠ ( z² + 1/z² ) = ( 6² + 2 )

➠ ( z² + 1/z² ) = 36+2

➠ ( z² + 1/z² ) = 38 ...........(2)

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Again,

★ ( a + b )² = ( a² + b² + 2ab )

so,

➠ ( z + 1/z )² = ( z² + 1/z² + 2 )

( keep value by (2) )

➠ ( z + 1/z )² = ( 38 + 2 )

➠ ( z + 1/z)² = (40)

➠ ( z +1/z) = √40 [ Ans.]

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