Math, asked by NewBornTigerYT, 9 months ago

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Answered by rishu6845
1

Answer:

\boxed{\huge{\pink{0 -  \dfrac{3}{2} i}}}

Step-by-step explanation:

\bold{\red{\underline{Given}}}\longrightarrow \\ real \: parts \: of \sqrt{5 + 2i}  \: and \:  \sqrt{5 - 2i}  \: are \: negativ

\bold{\orange{\underline{To \: find}}} \\ value \: of \\  \dfrac{  \sqrt{5 + 12i}   +  \sqrt{5 - 12i} }{ \sqrt{5 + 12i}  -  \sqrt{5 - 12i} }

\bold{\blue{\underline{Concept \: used }}}\longrightarrow\\  {i}^{2}  =  - 1

\bold{\green{\underline{Solution}}}\longrightarrow \\ let \\  \sqrt{5 + 12i}  = x + iy \\ squaring \: both \: sides \: we \: get

 {( \sqrt{5 + 12i}) }^{2}  =  {(x + iy)}^{2}  \\  =  > 5 + 12i =  {x}^{2}  +  {i}^{2}  {y}^{2}  + 2ixy \\  =  > 5 + 12i =  {x}^{2}  - 1 \:  {y}^{2}  + 2ixy \\  =  > 5 + 12i = ( {x}^{2}  -  {y}^{2} ) + (2xy)i \\ comparing \: real \: and \: imaginary \: part \: we \: get

 {x}^{2}  -  {y}^{2}  = 5.........(1) \\ 2xy = 12........(2)

now \: we \: know \: that \\  {( {x}^{2}  +  {y}^{2}) }^{2}  =  {( {x -  {y}^{2} )}^{2} } +  {(2xy)}^{2}   \\  =   {(5)}^{2}  +  {(12)}^{2}  \\  = 25 + 144 \\  =  >  {( {x}^{2} +  {y}^{2} ) }^{2}  = 169 \\ taking \: square \: root \: of \: both \: sides \\ and \: taking \:  +  \: sign \: because \: sum \:  \\ of \: squares \: can \: not \: negative \\  =  >  {x}^{2}  +  {y}^{2}  =   13......(3) \\ now \: adding \: (1) \: and \: (3) \: we \: get \\  {x}^{2}  +  {y}^{2}  +  {x}^{2}  -  {y}^{2}  = 13 + 5 \\  =  > 2 {x}^{2}  = 18 \\  =  >  {x}^{2}  = 9 \\ x = 3 \: and \: x =  - 3 \\ but \: real \: part \: of \:  \sqrt{5 + 12i} \\   \: is \: negative \: so \\ x =  - 3 \\ now \: putting \: x =  - 3 \: in \: (2) \\ 2xy = 12 \\  =  > 2( - 3)y = 12 \\  =  > y =  \dfrac{12}{ - 6}  \\  =  > y =  - 2 \\ so \\  \sqrt{5 + 12i}  =  - 3 - 2i

similarly  \\ \sqrt{5 - 12i}  =  - 3 + 12i

now

 \dfrac{ \sqrt{5 + 12i} +  \sqrt{5 - 12i}  }{ \sqrt{5 + 12i}  -  \sqrt{5 - 12i} }   \\  =  \dfrac{( - 3 - 2i) + ( - 3 + 2i}{( - 3 - 2i) - ( - 3 + 2i)}  \\  =  \dfrac{ - 3 - 2i - 3 + 2i}{ - 3 - 2i + 3 - 2i}  \\  =  \dfrac{ - 6}{ - 4i}  \\  =  \dfrac{3}{2i}  \\  =  \dfrac{3 \: i}{2  \: {i}^{2} }  \\  =  \dfrac{3 \: i}{2 \: ( - 1)}  \\  =  -  \dfrac{3}{2} i

 = 0 -  \dfrac{3}{2} i

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