Question

if a-ib = 3-4i/3+4i, then find a²+b²​

Answer 1

$\huge{\underline{\bf\red{QuestiØn} :}}$

$\sf If \: \boxed{a - ib = \frac{3 - 4i}{3 + 4i} } , \\ \sf then \: find \: {a}^{2} + {b}^{2} .$

$\huge{\underline{\bf\blue{SolutiØn}\ :}}$

We have,

$\begin{aligned} \\ \quad a - ib & = \frac{3 - 4i}{3 + 4i} \\ \\ & = \frac{3 - 4i}{3 + 4i} \times \frac{3 - 4i}{3 - 4i} \\ \\ & = \frac{ {(3 - 4i)}^{2} }{(3 - 4i)(3 + 4i)} \\ \\ & = \frac{ {(3)}^{2} + {(4i)}^{2} - 2(3)(4i) }{ {(3)}^{2} - {(4i)}^{2} } \\ \\ & = \frac{9 + 16( {i}^{2} ) - 24i}{9 - 16( {i}^{2} )} \\ \\ & = \frac{9 + 16( - 1) - 24i}{9 - 16( - 1)} \\ \\ & = \frac{9 - 16 - 24i}{9 + 16} \\ \\ & = \frac{ - 7 - 24i}{25} \\ \\ \implies a - ib & = \bigg( - \frac{7}{25} \bigg) - \bigg( \frac{24}{25} \bigg)i \end{aligned}$

On comparing both the sides, we get :

• $\boxed{a = - \frac{7}{25} }$

• $\boxed{b = \ \: \: \ \frac{24}{25} }$

HENCE,

$\begin{aligned} \\ a^2 + b^2 & = { \bigg( - \dfrac{7}{25} \bigg) }^{2} + { \bigg( \dfrac{24}{25} \bigg) }^{2} \\ \\ & = \dfrac{49}{625} + \dfrac{576}{625} \\ \\ & = \dfrac{49 + 576}{625} \\ \\ & = \dfrac{625}{625} \\ \\ & = \cancel{ \dfrac{625}{625} } \\ \\ & = 1 & & & & \end{aligned}$

$\red{\rule{5.5cm}{0.02cm}}$

$\Large{ \mid {\underline{\underline{\bf\green{BrainLiest \ AnswEr}}}} \mid }$