Complex problems

JEE Main 2015 Guidance › Algebra questions › Complex problems

Q1 FInd number of values of z for which e^z=0
Q2 Number of values of z for which sinz=0
Q3 Number of values of z for which cosz=0
Q4 What is solution set of sin(iy)=0

#Mathematics #Algebra

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9 Answers

106

Asish Mahapatra ·2009-03-24 00:51:18

e^z = 0
for no value i think

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106

Asish Mahapatra ·2009-03-24 00:52:12

sinz=0 ==> for sin to be defined z must be purely real and it is z=npi

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106

Asish Mahapatra ·2009-03-24 00:52:52

cosz=0 ==> again z is purely real and is equal to (2n+1)pi/2

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106

Asish Mahapatra ·2009-03-24 00:54:43

sin(iy) = 0 for sin function to be defined .. iy must be purely real
==> y is purely imaginary let = ix

So. it is sin(-x)=0
==> x=npi
==> y=i*npi

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1

Pavithra Ramamoorthy ·2009-03-24 01:15:36

fr second one it must be purely img ... isnt it???????? u r wron it seems......

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106

Asish Mahapatra ·2009-03-24 01:17:23

y?

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Pavithra Ramamoorthy ·2009-03-24 01:18:38

u temme hw do u say dat it must b purely real??????????

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106

Asish Mahapatra ·2009-03-24 01:22:22

arey sin function is defined for real numbers only.. what is the domain of sin function? thats why...

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1

Pavithra Ramamoorthy ·2009-03-24 01:27:23

k.. k..k..

srry.................

was jus thinkin it blindly.......

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