The expression tan i log e 2-3 i -2-3 i is equal to A- -12-3 B- -3-2 12 C- -12-5D- 0

Let 2+ 3i =√(13)e^i θ,tanθ= 3 2 ⇒ 2 3i = √(13)e^ i θ ⇒2 3i2 + 3i = e^ i2 θ nbsp; ⇒log_e ( 2 3i2 + 3i ) = i 2 θ ∴tan ( ilog_e 2 3i2 + 3i ) = tan2θ = ...

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Standard XII

Mathematics

Logarithm of a Complex Number

The expressio...

Question

The expression $tan (i {log}_{e} (\frac{2 - 3 i}{2 + 3 i}))$ is equal to

- \frac{12}{3}

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- \frac{3}{2}

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- \frac{12}{5}

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0

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Solution

The correct option is C $- \frac{12}{5}$
Let $2 + 3 i = \sqrt{13} e^{i θ}, tan θ = \frac{3}{2}$
$\Rightarrow 2 - 3 i = \sqrt{13} e^{- i θ}$
$\Rightarrow \frac{2 - 3 i}{2 + 3 i} = e^{- i 2 θ}$
$\Rightarrow {log}_{e} (\frac{2 - 3 i}{2 + 3 i}) = - i 2 θ$
$∴ tan (i {log}_{e} \frac{2 - 3 i}{2 + 3 i}) = tan 2 θ$
$= \frac{2 tan θ}{1 - {tan}^{2} θ}$
$= \frac{\frac{2 (3)}{2}}{1 - \frac{3^{2}}{2^{2}}}$
$= - \frac{12}{5}$

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The expression tan(iloge(2−3i2+3i)) is equal to

The correct option is C −125Let 2+3i=√13eiθ,tanθ=32 ⇒2−3i=√13e−iθ ⇒2−3i2+3i=e−i2θ ⇒loge(2−3i2+3i)=−i2θ ∴tan(iloge2−3i2+3i)=tan2θ =2tanθ1−tan2θ =2(3)21−3222 =−125

The expression $tan (i {log}_{e} (\frac{2 - 3 i}{2 + 3 i}))$ is equal to