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Answer 1

Answer:

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

Step-by-step explanation:

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

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

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

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

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

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

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

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

$now$

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

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