If coslog i4 i-a-b i- thenA- a-1- b -1B- a -1- b -2C- a -1- b -1D- a -1- b -0

The correct option is D a=1,b=0Given nbsp; cos(log i^4i)=a+bi ⇒ a+bi=cos(4ilog i) i = e^iπ /2⇒log i = nbsp; iπ2 So, a + bi = cos( 4i × nbsp; iπ2) ⇒ a + ...

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Standard XII

Mathematics

Complex Numbers

If coslog i4 ...

Question

If $cos (log i^{4 i}) = a + b i$ , then

a = 1, b = 2

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a = 1, b = - 1

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a = - 1, b = 1

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a = 1, b = 0

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Solution

The correct option is D $a = 1, b = 0$
Given
$cos (log i^{4 i}) = a + b i \Rightarrow a + b i = cos (4 i log i)$

$i = e^{i π / 2} \Rightarrow log i = \frac{i π}{2}$
So, $a + b i = cos (4 i \times \frac{i π}{2})$
$\Rightarrow a + b i = 1$
Hence, $a = 1, b = 0$

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If cos(logi4i)=a+bi, then

The correct option is D a=1,b=0Given cos(logi4i)=a+bi⇒a+bi=cos(4ilogi) i=eiπ/2⇒logi=iπ2 So, a+bi=cos(4i×iπ2) ⇒a+bi=1 Hence, a=1, b=0

If $cos (log i^{4 i}) = a + b i$ , then

The correct option is D $a = 1, b = 0$
Given
$cos (log i^{4 i}) = a + b i \Rightarrow a + b i = cos (4 i log i)$

$i = e^{i π / 2} \Rightarrow log i = \frac{i π}{2}$
So, $a + b i = cos (4 i \times \frac{i π}{2})$
$\Rightarrow a + b i = 1$
Hence, $a = 1, b = 0$