3-2i-2-5i-3-2i-2-5i is-

The correct option is B purely real 3+2i2 5i+3 2i2+5i =(3+2i)(2+5i)+(3 2i)(2 5i)(2 5i)(2+5i) = 4+19i+( 4 19i)4+25 = 829 Hence it is purely r ...

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Byju's Answer

Standard XII

Mathematics

Differentiation under Integral Sign

3+2i/2-5i+3-2...

Question

$\frac{3 + 2 i}{2 - 5 i} + \frac{3 - 2 i}{2 + 5 i}$ is:

purely real

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purely imaginary

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equal to zero

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2 + 5 i

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Solution

The correct option is B purely real
$\frac{3 + 2 i}{2 - 5 i} + \frac{3 - 2 i}{2 + 5 i}$
$= \frac{(3 + 2 i) (2 + 5 i) + (3 - 2 i) (2 - 5 i)}{(2 - 5 i) (2 + 5 i)}$
$= \frac{- 4 + 19 i + (- 4 - 19 i)}{4 + 25}$
$= \frac{- 8}{29}$
Hence it is purely real.

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Similar questions

\frac{2 + 5 i}{3 - 2 i} + \frac{2 - 5 i}{3 + 2 i}

Q. Simplify :

\frac{3 + 2 i}{2 - 5 i} + \frac{3 - 2 i}{2 + 5 i}

Imaginary part of third element of the second row for the Conjugate of $⎡ ⎢ ⎣ \begin{matrix} 3 + 4 i & 4 & 2 + 5 i 1 + 2 i & 2 + 3 i & 3 + 5 i 2 + 7 i & 9 & 5 \end{matrix} ⎤ ⎥ ⎦$ is

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Q. Solve for

z : z^{2} - (3 - 2 i) z = (5 i - 5)

Q. Let

z = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 + 2 i & - 5 i 1 - 2 i & - 3 & 5 + 3 i 5 i & 5 - 3 i & 7 \end{matrix} ∣ ∣ ∣ ∣

, then

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3+2i2−5i+3−2i2+5i is:

The correct option is B purely real3+2i2−5i+3−2i2+5i=(3+2i)(2+5i)+(3−2i)(2−5i)(2−5i)(2+5i)=−4+19i+(−4−19i)4+25=−829Hence it is purely real.

$\frac{3 + 2 i}{2 - 5 i} + \frac{3 - 2 i}{2 + 5 i}$ is:

The correct option is B purely real
$\frac{3 + 2 i}{2 - 5 i} + \frac{3 - 2 i}{2 + 5 i}$
$= \frac{(3 + 2 i) (2 + 5 i) + (3 - 2 i) (2 - 5 i)}{(2 - 5 i) (2 + 5 i)}$
$= \frac{- 4 + 19 i + (- 4 - 19 i)}{4 + 25}$
$= \frac{- 8}{29}$
Hence it is purely real.