Question

z=2-3i show that z²- 4z + 13 equal to zero

Answer 1

$\red{ \huge{ \underline{ \overline{ \mid{\bold{ \: \: SOLUTION \: \: \mid}}}}}}$

$\bold{\underline{Given\: \: : }\: \: \: \: \: \: \: \: \: z = 2 - 3i}$

$\bold{ \underline{To \: \: Prove \: \: :} \: \: \: \: \: \: z {}^{2} - 4z + 13 = 0}$



$\bold{Here,} \\ \\ \bold{z {}^{2} = (2 - 3i) {}^{2} } \\ \\ \bold{ = (2) {}^{2} - 2 \times 2 \times 3i + (3i) {}^{2} } \\ \\ \bold{ = 4 - 12i + 9i {}^{2} } \\ \\ \bold{ = 4 - 12i + 9( - 1) \: \: \: \: \: \: \: \: (as \: \: i {}^{2} = - 1) } \\ \\ \bold{ = 4 - 12i - 9} \\ \\ \bold{ = - 5 - 12i}$



$\bold{Again,} \\ \\ \bold{4z = 4(2 - 3i) = 8 - 12i}$



$\underline{\bold{ \: Putting \: \: the \: \: above \: \: values \: \: in \: \: L.H.S.}}$
$\underline{ \bold{ \: \: of\: \: the \: \: equation \: \: }}$


$\bold{L.H.S. = z {}^{2} - 4z + 13 } \\ \\ \bold{ = ( - 5 - 12i) - (8 - 12i) + 13} \\ \\ \bold{ = - 5 - 12i - 8 + 12i + 13} \\ \\ \bold{ = -1 3 + 13 -1 2i + 12i} \\ \\ \bold{ = 0} \\ \\ \bold{ = R.H.S.} \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bold{ \underline{ \: \: Hence \: \: Proved \: \: }}$

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✝$\mathfrak{ \red{THANKS..}}$✝

Answer 2

$\red{ \huge{ \underline{ \overline{ \mid{\bold{ \: \: SOLUTION \: \: \mid}}}}}}$

$\bold{\underline{Given\: \: : }\: \: \: \: \: \: \: \: \: z = 2 - 3i}$

$\bold{ \underline{To \: \: Prove \: \: :} \: \: \: \: \: \: z {}^{2} - 4z + 13 = 0}$

$\begin{lgathered}\bold{Here,} \\ \\ \bold{z {}^{2} = (2 - 3i) {}^{2} } \\ \\ \bold{ = (2) {}^{2} - 2 \times 2 \times 3i + (3i) {}^{2} } \\ \\ \bold{ = 4 - 12i + 9i {}^{2} } \\ \\ \bold{ = 4 - 12i + 9( - 1) \: \: \: \: \: \: \: \: (as \: \: i {}^{2} = - 1) } \\ \\ \bold{ = 4 - 12i - 9} \\ \\ \bold{ = - 5 - 12i}\end{lgathered}$

\begin{lgathered}\bold{Again,} \\ \\ \bold{4z = 4(2 - 3i) = 8 - 12i}\end{lgathered}

Again,

4z=4(2−3i)=8−12i

$\underline{\bold{ \: Putting \: \: the \: \: above \: \: values \: \: in \: \: L.H.S.}}$

$\underline{ \bold{ \: \: of\: \: the \: \: equation \: \: }}$

$\begin{lgathered}\bold{L.H.S. = z {}^{2} - 4z + 13 } \\ \\ \bold{ = ( - 5 - 12i) - (8 - 12i) + 13} \\ \\ \bold{ = - 5 - 12i - 8 + 12i + 13} \\ \\ \bold{ = -1 3 + 13 -1 2i + 12i} \\ \\ \bold{ = 0} \\ \\ \bold{ = R.H.S.} \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bold{ \underline{ \: \: Hence \: \: Proved \: \: }}\end{lgathered}$