For z - x-iy- x-y- R- define ez - ex cos y - isin y- sin z - 1-2ieiz - e-izand cos z - 1-2eiz - e-izNumber of values of z for which cos z - 0 is

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Theorems for Differentiability

For z = x+i...

Question

For
$z = x + i y, x, y \in R$ , define
$e^{z} = e^{x} (cos y + i sin y)$ ,
$sin z = \frac{1}{2 i} (e^{i z} - e^{- i z})$
and $cos z = \frac{1}{2} (e^{i z} + e^{- i z})$
Number of values of $z$ for which $cos z = 0$ is

0

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Infinite

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Solution

The correct option is D Infinite
For $c o s z = 0$

$e^{i z} + e^{- i z} = 0$

$e^{i z} = - e^{- i z}$

$e^{i 2 z} = - 1$

$e^{i 2 z} = e^{i (2 k - 1) π}$ ...( $k ϵ N$ )

$2 z = (2 k - 1) π$

$z = \frac{(2 k - 1) π}{2}$
Hence, option 'D' is correct.

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For z=x+iy,x,y∈R, defineez=ex(cosy+isiny),sinz=12i(eiz−e−iz)and cosz=12(eiz+e−iz)Number of values of z for which cosz=0 is

The correct option is D InfiniteFor cosz=0eiz+e−iz=0eiz=−e−izei2z=−1ei2z=ei(2k−1)π ...(kϵN)2z=(2k−1)πz=(2k−1)π2Hence, option 'D' is correct.

For
$z = x + i y, x, y \in R$ , define
$e^{z} = e^{x} (cos y + i sin y)$ ,
$sin z = \frac{1}{2 i} (e^{i z} - e^{- i z})$
and $cos z = \frac{1}{2} (e^{i z} + e^{- i z})$
Number of values of $z$ for which $cos z = 0$ is

The correct option is D Infinite
For $c o s z = 0$

$e^{i z} + e^{- i z} = 0$

$e^{i z} = - e^{- i z}$

$e^{i 2 z} = - 1$

$e^{i 2 z} = e^{i (2 k - 1) π}$ ...( $k ϵ N$ )

$2 z = (2 k - 1) π$

$z = \frac{(2 k - 1) π}{2}$
Hence, option 'D' is correct.