Question

1/3+4i+1 /3-4i(1 divided by 3+4i plus 1 divided by 3-4i)​

Answer 1

Answer:

$\sf{The \ value \ of \ \dfrac{1}{3+4i}+\dfrac{1}{3-4i}}$

$\sf{is \ \dfrac{6}{25}.}$

To find:

$\sf{The \ value \ of \ \dfrac{1}{3+4i}+\dfrac{1}{3-4i}}$

Solution:

$\sf{\leadsto{\dfrac{1}{3+4i}+\dfrac{1}{3-4i}}}$

$\sf{\leadsto{\dfrac{3-4i+3+4i}{(3+4i)(3-4i)}}}$

$\sf{\leadsto{\dfrac{6}{9-16i^{2}}}}$

$\sf{\leadsto{\dfrac{6}{9+16}}}$

$\sf{\leadsto{\dfrac{6}{25}}}$

$\sf\purple{\therefore{\tt{The \ value \ of \ \dfrac{1}{3+4i}+\dfrac{1}{3-4i}}}}$

$\sf\purple{\tt{is \ \dfrac{6}{25}.}}$

____________________________

Extra information:

• $\sf{i=\sqrt{-1}}$

• $\sf{i^{2}=-1}$

• $\sf{i^{3}=-i}$

• $\sf{i^{4}=1}$

Answer 2

To Find :

• value of 1/3+4i+1 /3-4i

Solution :

$= \rm \large{ \frac{1}{3 + 4i} + \frac{1}{3 - 4i} }$

• Taking LCM

$= \rm \large{ \frac{(3 - 4i) + (3 + 4i)}{(3 + 4i).(3 - 4i)} }$

› By using identity (a + b) (a - b) = a² - b²

$= \rm \large{ \frac{3 - 4i + 3 + 4i}{9 - 16 {i}^{2} }}$

As we know that,

• i² = -1

So,

$= \rm \large{ \frac{6}{9 - 16 \times ( - 1)} }$

$= \rm \large{ \frac{6}{9 + 16} }$

$= \rm \large{ \frac{6}{25} }$

$\large\dag \large { \red{\underline{\bf{Hence }}}}$

$\green{\textrm{ Value of 1/(3+4i)+1 /(3-4i) is 6/25 }}$

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