z=π4 पर फलन f(z)=1coszsinz की अव्युत्क्रमणीयता की प्रकृति है

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ESE Electrical 2019 Official Paper
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  1. विस्थापनीय अव्युत्क्रमणीयता
  2. विलगित अव्युत्क्रमणीयता
  3. सरल ध्रुव
  4. अनिवार्य अव्युत्क्रमणीयता

Answer (Detailed Solution Below)

Option 3 : सरल ध्रुव
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संकल्पना:

1. z0(zz0)f(z)|z=z0=0 पर विस्थापनीय अव्युत्क्रमणीयता

2. z0(zz0)f(z)|zz0 = परिमित मान पर सरल ध्रुव

3. अनिवार्य अव्युत्क्रमणीयताएँ → बहुत ऊँची कोटि का ध्रुव (या अपरिमित)

गणना:

दिया गया है f(z)=1coszsinz=(cosz+sinz)cos2zsin2z

f(z)=(cosz+sinz)cos2z (चूँकि cos2 θ – sin2 θ = cos 2 θ)

z=t+π4 रखने पर इसलिए हम निम्न पर अव्युत्क्रमणीयता प्राप्त कर सकते हैं

z=π4 (t → 0)

f(z)|z=t+π4=g(t)1[2t(2t)33!+(2t)55!+]=g(t)(t)[2(2)3t23!+28t45!]

उपरोक्त से यह स्पष्ट है कि

f(z)|z=t+π4 का t = 0 पर एक सरल ध्रुव है।

इसका तात्पर्य यह है कि f(z) का z=t+π4=π4 पर सरल ध्रुव है
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