Perform the following operation and write the result in standard form.
$$\frac { 9 i } { 9 + i } + \frac { 2 } { 9 - i }$$

A) $\frac { 9 } { 40 } + \frac { 83 } { 80 } i$

B) $\frac { 27 } { 10 } + \frac { 83 } { 10 } i$

C) $\frac { 1 } { 41 } + \frac { 9 } { 82 } i$

D) $\frac { 27 } { 82 } + \frac { 83 } { 82 } i$

E) $\frac { 27 } { 8 } + \frac { 83 } { 8 } i$

Question

Perform the following operation and write the result in standard form.
$$\frac { 9 i } { 9 + i } + \frac { 2 } { 9 - i }$$

A) $\frac { 9 } { 40 } + \frac { 83 } { 80 } i$

B) $\frac { 27 } { 10 } + \frac { 83 } { 10 } i$

C) $\frac { 1 } { 41 } + \frac { 9 } { 82 } i$

D) $\frac { 27 } { 82 } + \frac { 83 } { 82 } i$

E) $\frac { 27 } { 8 } + \frac { 83 } { 8 } i$

Quizplus · Accepted Answer

To simplify the given expression, we first rationalize the denominators of both fractions by multiplying the numerator and denominator by the conjugate of the denominator. For the first fraction, the conjugate of $9 + i$ is $9 - i$, and for the second fraction, the conjugate of $9 - i$ is $9 + i$. Simplifying the first fraction:$$\frac{9i}{9+i} \cdot \frac{9-i}{9-i} = \frac{9i(9-i)}{81+1} = \frac{81i-9i^2}{82} = \frac{81i+9}{82} = \frac{9}{82} + \frac{81i}{82}$$Simplifying the second fraction:$$\frac{2}{9-i} \cdot \frac{9+i}{9+i} = \frac{2(9+i)}{81+1} = \frac{18+2i}{82} = \frac{18}{82} + \frac{2i}{82} = \frac{9}{41} + \frac{i}{41}$$Adding the simplified fractions:$$\frac{9}{82} + \frac{81i}{82} + \frac{9}{41} + \frac{i}{41} = \frac{9}{82} + \frac{18}{82} + \frac{81i}{82} + \frac{2i}{82} = \frac{27}{82} + \frac{83i}{82}$$Therefore, the result in standard form is $\frac{27}{82} + \frac{83i}{82}$, which matches choice D.

Perform the Following Operation and Write the Result in Standard 9i9+i+29−i\frac { 9 i } { 9 + i } + \frac { 2 } { 9 - i }9+i9i​+9−i2​

Perform the Following Operation and Write the Result in Standard $\frac { 9 i } { 9 + i } + \frac { 2 } { 9 - i }$