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C R Equations Prof. Anil Vinayakrao Patil

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A C S College Navapur Department Of Mathematics

𝑃𝑟𝑜𝑓. 𝐴𝑛𝑖𝑙 𝑉𝑖𝑛𝑎𝑦𝑎𝑘𝑟𝑎𝑜 𝑃𝑎𝑡𝑖𝑙

𝑪𝒂𝒖𝒄𝒉𝒚 − 𝑹𝒊𝒆𝒎𝒂𝒏𝒏 𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏𝒔:

𝐿𝑒𝑡 𝑢 𝑎𝑛𝑑 𝑣 𝑏𝑒 𝑟𝑒𝑎𝑙 𝑣𝑎𝑙𝑢𝑒𝑑 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠 𝑜𝑓 𝑡𝑤𝑜 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 𝑥 𝑎𝑛𝑑 𝑦 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 𝑢𝑥 = 𝑣𝑦 𝑎𝑛𝑑 𝑢𝑦 = −𝑣𝑥 𝑎𝑟𝑒 𝑐𝑎𝑙𝑙𝑒𝑑 𝐶𝑎𝑢𝑐ℎ𝑦 − 𝑅𝑖𝑒𝑚𝑎𝑛𝑛 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 (𝑖. 𝑒. 𝐶 − 𝑅 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠)

𝑇ℎ𝑒𝑜𝑟𝑒𝑚 ∶ 𝐴 𝑛𝑒𝑐𝑒𝑠𝑠𝑎𝑟𝑦 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 𝑡ℎ𝑎𝑡 𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑤 = 𝑓 𝑧 = 𝑢 𝑥, 𝑦 + 𝑖𝑣 𝑥, 𝑦 𝑏𝑒 𝑎𝑛𝑎𝑙𝑦𝑡𝑖𝑐 𝑎𝑡 𝑝𝑜𝑖𝑛𝑡 𝑧 = 𝑥 + 𝑖𝑦 𝑖𝑠 𝑡ℎ𝑎𝑡 𝑎𝑡 𝑝𝑜𝑖𝑛𝑡 𝑥, 𝑦 𝑟𝑒𝑎𝑙 𝑎𝑛𝑑 𝑖𝑚𝑎𝑧𝑖𝑛𝑎𝑟𝑦 𝑝𝑎𝑟𝑡𝑠 𝑢 𝑎𝑛𝑑 𝑣 𝑜𝑓 𝑓 𝑧 𝑠𝑎𝑡𝑖𝑠𝑓𝑖𝑒𝑠 𝑡ℎ𝑒 𝐶 − 𝑅 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠. 𝑃𝑟𝑜𝑜𝑓 ∶

𝐿𝑒𝑡 𝑤 = 𝑓 𝑧 = 𝑢 𝑥, 𝑦 + 𝑖𝑣 𝑥, 𝑦 𝑏𝑒 𝑎𝑛𝑎𝑙𝑦𝑡𝑖𝑐 𝑎𝑡 𝑝𝑜𝑖𝑛𝑡 𝑧 = 𝑥 + 𝑖𝑦

𝑓 𝑥+𝛿𝑥+𝑖𝑦 −𝑓(𝑥+𝑖𝑦) 𝛿𝑥 𝛿𝑥→0

∴ 𝑓 ′ 𝑧 𝑖𝑠 𝑒𝑥𝑖𝑠𝑡𝑠

∴ 𝑓 ′ 𝑧 = lim

𝑓 𝑧+𝛿𝑧 −𝑓(𝑧) 𝛿𝑧 𝛿𝑧→0

∴ 𝑓 ′ 𝑧 = lim

𝑖𝑠 𝑒𝑥𝑖𝑠𝑡𝑠 … … . . (𝐼)

∴ 𝐴𝑙𝑜𝑛𝑔 𝑑𝑖𝑓𝑓𝑟𝑒𝑛𝑡 𝑝𝑎𝑡ℎ𝑠 𝑡ℎ𝑖𝑠 𝑙𝑖𝑚𝑖𝑡 𝑖𝑠 𝑠𝑎𝑚𝑒.

𝐼) 𝑙𝑖𝑚𝑖𝑡 𝑎𝑙𝑜𝑛𝑔 𝑟𝑒𝑎𝑙 𝑎𝑥𝑖𝑠 𝑖. 𝑒. 𝑋 − 𝑎𝑥𝑖𝑠 : 𝐴𝑠 𝑧 = 𝑥 + 𝑖𝑦 ∴ 𝛿𝑧 = 𝛿𝑥 + 𝑖𝛿𝑦 ∴ 𝛿𝑧 = 𝛿𝑥 + 𝑖. 0 = 𝛿𝑥 ∴ 𝑏𝑦 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝐼 𝑓 𝑥+𝑖𝑦+𝛿𝑥 −𝑓(𝑥+𝑖𝑦) 𝛿𝑥 𝛿𝑥→0

= lim

𝛿𝑥→0

= lim

𝛿𝑥→0

𝐻𝑒𝑟𝑒 𝑤𝑒 ℎ𝑎𝑣𝑒 𝑡𝑜 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑡𝑤𝑜 𝑝𝑎𝑡ℎ𝑠, 𝑎𝑙𝑜𝑛𝑔 𝑟𝑒𝑎𝑙 𝑎𝑥𝑖𝑠 𝑎𝑛𝑑 𝑖𝑚𝑎𝑧𝑖𝑛𝑎𝑟𝑦 𝑎𝑥𝑖𝑠

𝑓 ′ 𝑧 = lim

As 𝑓 𝑥 + 𝑖𝑦 = 𝑢 𝑥, 𝑦 + 𝑖𝑣 𝑥, 𝑦

= lim

𝛿𝑥→0

𝑢 𝑥+𝛿𝑥,𝑦 +𝑖𝑣 𝑥+𝛿𝑥,𝑦 − 𝑢 𝑥,𝑦 +𝑖𝑣 𝑥,𝑦 𝛿𝑥 𝑢 𝑥+𝛿𝑥,𝑦 −𝑢(𝑥,𝑦) +𝑖 𝑣 𝑥+𝛿𝑥,𝑦 −𝑣(𝑥,𝑦) 𝛿𝑥

𝑢 𝑥+𝛿𝑥,𝑦 −𝑢(𝑥,𝑦) 𝛿𝑥

𝑢 𝑥+𝛿𝑥,𝑦 −𝑢(𝑥,𝑦) 𝛿𝑥 𝛿𝑥→0

= lim

+𝑖

𝑣 𝑥+𝛿𝑥,𝑦 −𝑣(𝑥,𝑦) 𝛿𝑥 𝑣 𝑥+𝛿𝑥,𝑦 −𝑣(𝑥,𝑦) 𝛿𝑥 𝛿𝑥→0

+ 𝑖 lim

∴ 𝑓 ′ 𝑧 = 𝑢𝑥 𝑥, 𝑦 + 𝑖 𝑣𝑥 𝑥, 𝑦 … … … (𝐼𝐼) 𝐼𝐼) 𝑙𝑖𝑚𝑖𝑡 𝑎𝑙𝑜𝑛𝑔 𝑖𝑚𝑎𝑧𝑖𝑛𝑎𝑟𝑦 𝑎𝑥𝑖𝑠 𝑖. 𝑒. 𝑌 − 𝑎𝑥𝑖𝑠 : 𝐴𝑠 𝛿𝑧 = 𝛿𝑥 + 𝑖𝛿𝑦

∴ 𝛿𝑧 = 0 + 𝑖𝛿𝑦

∴ 𝛿𝑧 = 𝑖𝛿𝑦

∴ 𝑏𝑦 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝐼 𝑓 𝑥+𝑖𝑦+𝑖𝛿𝑦 −𝑓(𝑥+𝑖𝑦) 𝑖𝛿𝑦 𝑖𝛿𝑦→0

𝑓 ′ 𝑧 = lim

𝑓 𝑥+𝑖𝑦+𝑖𝛿𝑦 −𝑓(𝑥+𝑖𝑦) 𝑖𝛿𝑦 𝑖𝛿𝑦→0

∴ 𝑓 ′ 𝑧 = lim

∴ 𝑓 ′ 𝑧 = −𝑖𝑢𝑦 𝑥, 𝑦 + 𝑣𝑦 𝑥, 𝑦

𝑓 𝑥+𝑖(𝑦+𝛿𝑦) −𝑓(𝑥+𝑖𝑦) 𝑖𝛿𝑦 𝛿𝑦→0

∴ 𝑓 ′ 𝑧 = 𝑣𝑦 𝑥, 𝑦 − 𝑖𝑢𝑦 𝑥, 𝑦

1

∴ 𝐵𝑦 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 𝐼𝐼 𝑎𝑛𝑑 𝐼𝐼𝐼 𝑤𝑒 ℎ𝑎𝑣𝑒

= lim

𝑓 𝑥+𝑖(𝑦+𝛿𝑦) −𝑓(𝑥+𝑖𝑦) 𝛿𝑦 𝛿𝑦→0

= 𝑖 lim

=

∴ 𝑢𝑥 𝑥, 𝑦 + 𝑖 𝑣𝑥 𝑥, 𝑦 = 𝑣𝑦 𝑥, 𝑦 − 𝑖𝑢𝑦 𝑥, 𝑦

1 lim 𝑖 𝛿𝑦→0

𝑢 𝑥,𝑦+𝛿𝑦 +𝑖𝑣 𝑥,𝑦+𝛿𝑦 − 𝑢 𝑥,𝑦 +𝑖𝑣 𝑥,𝑦 𝛿𝑦

1

𝑢 𝑥,𝑦+𝛿𝑦 −𝑢(𝑥,𝑦) +𝑖 𝑣 𝑥,𝑦+𝛿𝑦 −𝑣(𝑥,𝑦) 𝛿𝑦

= 𝑖 lim

𝛿𝑦→0

… … … … . . (𝐼𝐼𝐼)

∴ 𝑢𝑥 𝑥, 𝑦 + 𝑖 𝑣𝑥 𝑥, 𝑦 = 𝑣𝑦 𝑥, 𝑦 + 𝑖(−𝑢𝑦 𝑥, 𝑦 ) 𝐶𝑜𝑚𝑝𝑎𝑟𝑖𝑛𝑔 𝑟𝑒𝑎𝑙 𝑎𝑛𝑑 𝑖𝑚𝑎𝑧𝑖𝑛𝑎𝑟𝑦 𝑝𝑎𝑟𝑡𝑠 𝑜𝑛 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠, 𝑤𝑒 𝑔𝑒𝑡 𝑢𝑥 𝑥, 𝑦 = 𝑣𝑦 𝑥, 𝑦 𝑎𝑛𝑑 𝑣𝑥 𝑥, 𝑦 = −𝑢𝑦 𝑥, 𝑦

1

= 𝑖 lim

𝛿𝑦→0

1

𝑢 𝑥,𝑦+𝛿𝑦 −𝑢(𝑥,𝑦) 𝛿𝑦

𝑢 𝑥,𝑦+𝛿𝑦 −𝑢(𝑥,𝑦) 𝛿𝑦 𝛿𝑦→0

= 𝑖 lim

+𝑖

∴ 𝑢𝑥 𝑥, 𝑦 = 𝑣𝑦 𝑥, 𝑦 𝑎𝑛𝑑 𝑢𝑦 𝑥, 𝑦 = − 𝑣𝑥 𝑥, 𝑦

𝑣 𝑥,𝑦+𝛿𝑦 −𝑣(𝑥,𝑦) 𝛿𝑦

𝑣 𝑥,𝑦+𝛿𝑦 −𝑣(𝑥,𝑦) 𝛿𝑦 𝛿𝑦→0

+ 𝑖 lim

∴ 𝑅𝑒𝑎𝑙 𝑎𝑛𝑑 𝑖𝑚𝑎𝑧𝑖𝑛𝑎𝑟𝑦 𝑝𝑎𝑟𝑡𝑠 𝑢 𝑎𝑛𝑑 𝑣 𝑜𝑓 𝑎𝑛𝑎𝑙𝑦𝑡𝑖𝑐𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑓 𝑧 𝑠𝑎𝑡𝑖𝑠𝑓𝑖𝑒𝑠 𝑡ℎ𝑒 𝐶 − 𝑅 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 𝑎𝑡 𝑝𝑜𝑖𝑛𝑡 𝑥, 𝑦 .

1

= 𝑖 𝑢𝑦 𝑥, 𝑦 + 𝑖 𝑣𝑦 𝑥, 𝑦 ∴ 𝑓 ′ 𝑧 = − 𝑖 𝑢𝑦 𝑥, 𝑦 + 𝑖 𝑣𝑦 𝑥, 𝑦

∴ 𝑰𝒇 𝒂 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏 𝒇 𝒛 = 𝒖 + 𝒊𝒗 𝒃𝒆 𝒂𝒏𝒂𝒍𝒚𝒕𝒊𝒄 𝒂𝒕 𝒑𝒐𝒊𝒏𝒕 𝒛 = 𝒙 + 𝒊𝒚 𝒕𝒉𝒆𝒏 𝒖 𝒂𝒏𝒅 𝒗 𝒐𝒇 𝒔𝒂𝒕𝒊𝒔𝒇𝒊𝒆𝒔 𝒕𝒉𝒆 𝑪 − 𝑹 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏𝒔. 𝐻𝑒𝑛𝑐𝑒 𝑡ℎ𝑒 𝑝𝑟𝑜𝑜𝑓 𝑜𝑓 𝑡ℎ𝑒 𝑡ℎ𝑒𝑜𝑟𝑒𝑚. 𝑹𝒆𝒎𝒂𝒓𝒌 ∶ 𝑇ℎ𝑒 𝑐𝑜𝑛𝑣𝑒𝑟𝑠𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑡ℎ𝑒𝑜𝑟𝑚 𝑖𝑠 𝑛𝑜𝑡 𝑡𝑟𝑢𝑒.

𝑬𝒙𝒂𝒎𝒑𝒍𝒆𝒔 I)

𝑆ℎ𝑜𝑤 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑓 𝑧 = 𝑥𝑦 𝑖𝑠 𝑛𝑜𝑡 𝑎𝑛𝑎𝑙𝑦𝑡𝑖𝑐 𝑎𝑡 𝑜𝑟𝑖𝑔𝑖𝑛 𝑡ℎ𝑜𝑢𝑔ℎ 𝑡ℎ𝑒 𝐶. 𝑅. 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 𝑎𝑟𝑒 𝑠𝑎𝑡𝑖𝑠𝑓𝑖𝑒𝑑 𝑎𝑡 𝑡ℎ𝑎𝑡 𝑝𝑜𝑖𝑛𝑡. 𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏 ∶ 𝐹𝑖𝑟𝑠𝑡 𝑡𝑜 𝑠ℎ𝑜𝑤 𝑟𝑒𝑎𝑙 𝑎𝑛𝑑 𝑖𝑚𝑎𝑧𝑖𝑛𝑎𝑟𝑦 𝑝𝑎𝑟𝑡𝑠 𝑜𝑓 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑓 𝑧 𝑠𝑎𝑡𝑖𝑒𝑠𝑓𝑖𝑒𝑠 𝐶. 𝑅. 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 𝑎𝑡 𝑜𝑟𝑖𝑔𝑖𝑛 𝑢 𝑥, 𝑦 + 𝑘 − 𝑢(𝑥, 𝑦) 𝐴𝑠 𝑓 𝑧 = 𝑥𝑦 𝐴𝑠 𝑢𝑦 𝑥, 𝑦 = lim 𝑘→0 𝑘 ∴ 𝑓 𝑧 = 𝑥𝑦 + 𝑖. 0 = 𝑢 𝑥, 𝑦 + 𝑖𝑣(𝑥, 𝑦)

𝑢 0, 𝑘 − 𝑢(0,0) ∴ 𝑢𝑦 0,0 = lim 𝑘→0 𝑘 ∴ 𝑢 𝑥, 𝑦 = 𝑥𝑦 & 𝑣 𝑥, 𝑦 = 0 0. 𝑘 − 0.0 𝑢 𝑥 + ℎ, 𝑦 − 𝑢(𝑥, 𝑦) ∴ 𝑢𝑦 0,0 = lim 𝐴𝑠 𝑢𝑥 𝑥, 𝑦 = lim 𝑘→0 𝑘 ℎ→0 ℎ 0−0 𝑢 ℎ, 0 − 𝑢(0,0) ∴ 𝑢𝑦 0,0 = lim 𝑘→0 𝑘 ∴ 𝑢𝑥 0,0 = lim ℎ→0 ℎ ∴ 𝑢𝑦 0,0 = lim 0 𝑘→0 ℎ. 0 − 0.0 ∴ 𝑢𝑥 0,0 = lim ∴ 𝑢𝑦 0,0 = 0 ℎ→0 ℎ 𝑁𝑜𝑤 𝑎𝑠 𝑣 𝑥, 𝑦 = 0 ∴ 𝑣𝑥 𝑥, 𝑦 = 0 𝑎𝑛𝑑 𝑣𝑦 𝑥, 𝑦 = 0 0−0 ∴ 𝑢𝑥 0,0 = lim ∴ 𝑣𝑥 0,0 = 0 𝑎𝑛𝑑 𝑣𝑦 0,0 = 0 ℎ→0 ℎ ∴ 𝑢𝑥 0,0 = lim 0 ℎ→0

∴ 𝑢𝑥 0,0 = 0

∴ 𝑢𝑥 0,0 = 𝑣𝑦 0,0

𝑎𝑛𝑑 𝑢𝑦 0,0 = − 𝑣𝑥 0,0

∴ 𝒖 𝒂𝒏𝒅 𝒗 𝒔𝒂𝒕𝒊𝒆𝒔𝒇𝒊𝒆𝒔 𝑪. 𝑹. 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏𝒔 𝒂𝒕 𝒐𝒓𝒊𝒈𝒊𝒏.

𝑁𝑜𝑤 𝑡𝑜 𝑠ℎ𝑜𝑤 𝑓 𝑧 𝑖𝑠 𝑛𝑜𝑡 𝑎𝑛𝑎𝑙𝑦𝑡𝑖𝑐 𝑎𝑡 𝑜𝑟𝑖𝑔𝑖𝑛. 𝑓 𝑧 − 𝑓(𝑧0 ) = lim z→𝑧0 𝑧 − 𝑧0

𝐴𝑠 𝑓 ′ 𝑧0

𝑓 𝑧 − 𝑓(0) 𝐴𝑠 𝑓 0 = lim z→0 𝑧−0 ′

𝐺𝑖𝑣𝑒𝑛 𝑡ℎ𝑎𝑡

𝑓 𝑧 =

∴ 𝑓 ′ 0 = lim

z→0

∴

𝑓′

lim

x+i.mx→0

∴ 𝑓 ′ 0 = lim

𝑚 (1 + 𝑖. 𝑚)

x→0

x→0

𝑚 (1 + 𝑖. 𝑚)

𝑤ℎ𝑖𝑐ℎ 𝑑𝑒𝑝𝑒𝑛𝑑𝑠 𝑜𝑛 𝑚, 𝑓𝑜𝑟 𝑑𝑖𝑓𝑓𝑟𝑒𝑛𝑡

𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑚 𝑤𝑒 ℎ𝑎𝑣𝑒 𝑑𝑖𝑓𝑓𝑟𝑒𝑛𝑡 𝑙𝑖𝑚𝑖𝑡𝑠 .

𝑥𝑦 − 0 𝑧

∴ 𝐹𝑜𝑟 𝑑𝑖𝑓𝑓𝑟𝑒𝑛𝑡 𝑝𝑎𝑡ℎ𝑠 𝑙𝑖𝑚𝑖𝑡𝑠 𝑎𝑟𝑒 𝑑𝑖𝑓𝑓𝑟𝑒𝑛𝑡, 𝑠𝑜 𝑙𝑖𝑚𝑖𝑡 𝑖𝑛 𝑡ℎ𝑒 𝑟𝑖𝑔ℎ𝑡 ℎ𝑎𝑛𝑑 𝑠𝑖𝑑𝑒 𝑜𝑓 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝐼 𝑑𝑜𝑒𝑠𝑛′ 𝑡 𝑒𝑥𝑖𝑠𝑡.

… … … …(I)

𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑡ℎ𝑒 𝑝𝑎𝑡ℎ 𝑦 = 𝑚𝑥 ∴ 𝑓′ 0 =

𝑥 𝑚 𝑥(1 + 𝑖. 𝑚)

∴ 𝑓′ 0 =

𝑥𝑦

𝑥𝑦 x+iy→0 𝑥+𝑖𝑦

∴ 𝑓 ′ 0 = lim

∴ 𝑓 ′ 0 = lim

𝑥. 𝑚𝑥 𝑥 + 𝑖. 𝑚𝑥

𝑥 2. 𝑚 0 = lim x→0 𝑥(1 + 𝑖. 𝑚)

∴ 𝑓 ′ 0 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑒𝑥𝑖𝑠𝑡𝑠. ∴ 𝑮𝒊𝒗𝒆𝒏 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏 𝒇 𝒛 𝒏𝒐𝒕 𝒂𝒏𝒂𝒍𝒚𝒕𝒊𝒄 𝒂𝒕 𝒐𝒓𝒊𝒈𝒊𝒏 𝒕𝒉𝒐𝒖𝒈𝒉 𝒕𝒉𝒆 𝑪. 𝑹. 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏𝒔 𝒂𝒓𝒆 𝒔𝒂𝒕𝒊𝒔𝒇𝒊𝒆𝒅 𝒂𝒕 𝒕𝒉𝒂𝒕 𝒑𝒐𝒊𝒏𝒕.

II)

𝐼𝑓 𝑓 𝑧

𝑥𝑦 2 (𝑥 + 𝑖𝑦) , = ൞ 𝑥2 + 𝑦4 0 ,

𝑖𝑓 𝑧 ≠ 0

ℎ→0

𝑖𝑓 𝑧 = 0

𝑆ℎ𝑜𝑤 𝑡ℎ𝑎𝑡 𝐶. 𝑅. 𝑒𝑞𝑢𝑞𝑡𝑖𝑜𝑛𝑠 𝑎𝑟𝑒 𝑠𝑎𝑡𝑖𝑠𝑓𝑖𝑒𝑠 𝑎𝑡 𝑧 = 0 𝑏𝑢𝑡 𝑓 ′ 0 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑒𝑥𝑖𝑠𝑡𝑠. 𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏 ∶ 𝐺𝑖𝑣𝑒𝑛 𝑓 𝑧

𝑥𝑦 2 (𝑥 + 𝑖𝑦) , = ൞ 𝑥2 + 𝑦4 0 ,

𝑖𝑓 𝑧 ≠ 0 𝑖𝑓 𝑧 = 0

𝑢 ℎ, 0 − 𝑢(0,0) ℎ

𝐴𝑠 𝑢𝑥 0,0 = lim

∴ 𝑢𝑥

ℎ2 . 0 2+0−0 ℎ 0,0 = lim ℎ→0 ℎ

∴ 𝑢𝑥

0−0 0,0 = lim ℎ→0 ℎ

∴ 𝑢𝑥 0,0 = lim 0 ℎ→0

𝑥𝑦 2 (𝑥 + 𝑖𝑦) 𝐴𝑠 𝑓 𝑧 = 𝑥2 + 𝑦4

∴ 𝑢𝑥 0,0 = 0 𝑢 0, 𝑘 − 𝑢(0,0) 𝑁𝑜𝑤 𝑢𝑦 0,0 = lim 𝑘→0 𝑘

𝑥 2 𝑦 2 + 𝑖𝑥𝑦 3 ∴ 𝑓 𝑧 = 𝑥2 + 𝑦4 𝑥 2𝑦2 𝑥𝑦 3 ∴ 𝑓 𝑧 = 2 +𝑖 2 𝑥 + 𝑦4 𝑥 + 𝑦4 = 𝑢 𝑥, 𝑦 + 𝑖𝑣(𝑥, 𝑦) ∴ 𝑢 𝑥, 𝑦 =

𝑥 2𝑦2 𝑥2 + 𝑦4

& 𝑣 𝑥, 𝑦 =

𝑥𝑦 3 𝑥2 + 𝑦4

0. 𝑘 2 4−0 0 + 𝑘 ∴ 𝑢𝑦 0,0 = lim 𝑘→0 𝑘 ∴ 𝑢𝑦 0,0 = lim

𝑘→0

∴ 𝑢𝑦 0,0 = 0

0−0 𝑘

𝑆𝑖𝑚𝑖𝑙𝑎𝑟𝑙𝑦

′

∴ 𝑓 0 =

𝑣𝑥 0,0 = 0 𝑎𝑛𝑑 𝑣𝑦 0,0 = 0

∴ 𝑢𝑥 0,0 = 𝑣𝑦 0,0

𝑎𝑛𝑑 𝑢𝑦 0,0 = − 𝑣𝑥 0,0

As

∴ 𝑢 𝑎𝑛𝑑 𝑣 𝑠𝑎𝑡𝑖𝑒𝑠𝑓𝑖𝑒𝑠 𝐶. 𝑅. 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠 𝑎𝑡 𝑜𝑟𝑖𝑔𝑖𝑛. 𝑁𝑜𝑤 𝑡𝑜 𝑠ℎ𝑜𝑤 𝑓 ′ 0 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑒𝑥𝑖𝑠𝑡𝑠 𝐴𝑠 𝑓 ′ 𝑧0

𝑓 𝑧 − 𝑓(𝑧0 ) = lim z→𝑧0 𝑧 − 𝑧0

∴ 𝑓 ′ 0 = lim

z→0

𝑓 𝑧 − 𝑓(0) 𝑧−0

x+iy→0

∴

𝑓′

0 =

𝑇𝑎𝑘𝑒 𝑡ℎ𝑒 𝑝𝑎𝑡ℎ 𝑦 2 = 𝑚𝑥 𝑖. 𝑒. 𝑦 = 𝑚𝑥

𝑚𝑥 2 0 = lim 2 x→0 𝑥 (1 + 𝑚 2 )

x→0

∴

+ 𝑖𝑦) −0 𝑥2 + 𝑦4 𝑥 + 𝑖𝑦

𝑥𝑦 2 lim x+iy→0 𝑥 2 +𝑦 4

∴

x + i 𝑚𝑥 → 0 ∴ x + i 𝑚𝑥 → 0 + i. 0 ⟹ x → 0 & 𝑚𝑥 → 0 ⟹x→0

∴ 𝑓 ′ 0 = lim

𝑥𝑦 2 (𝑥 ∴ 𝑓 ′ 0 = lim

𝑓′

lim

x+i 𝑚𝑥→0

……………(I)

𝑥. 𝑚𝑥 𝑥 2 + 𝑚2 𝑥 2

𝑓′

𝑚 (1 + 𝑚2 )

𝑚 0 = 𝑤ℎ𝑖𝑐ℎ 𝑑𝑒𝑝𝑒𝑛𝑑𝑠 𝑜𝑛 𝑚, 𝑓𝑜𝑟 𝑑𝑖𝑓𝑓𝑟𝑒𝑛𝑡 (1 + 𝑚2 ) 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑚 𝑤𝑒 ℎ𝑎𝑣𝑒 𝑑𝑖𝑓𝑓𝑟𝑒𝑛𝑡 𝑙𝑖𝑚𝑖𝑡𝑠.

∴ 𝐹𝑜𝑟 𝑑𝑖𝑓𝑓𝑟𝑒𝑛𝑡 𝑝𝑎𝑡ℎ𝑠 𝑙𝑖𝑚𝑖𝑡𝑠 𝑎𝑟𝑒 𝑑𝑖𝑓𝑓𝑟𝑒𝑛𝑡, 𝑠𝑜 𝑙𝑖𝑚𝑖𝑡 𝑖𝑛 𝑡ℎ𝑒 𝑟𝑖𝑔ℎ𝑡 ℎ𝑎𝑛𝑑 𝑠𝑖𝑑𝑒 𝑜𝑓 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝐼 𝑑𝑜𝑒𝑠𝑛′ 𝑡 𝑒𝑥𝑖𝑠𝑡𝑠. ∴ 𝑓 ′ 0 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑒𝑥𝑖𝑠𝑡𝑠.

∴ 𝑢 𝑎𝑛𝑑 𝑣 𝑠𝑎𝑡𝑖𝑠𝑓𝑖𝑒𝑠 𝐶. 𝑅. 𝑒𝑞𝑢𝑞𝑡𝑖𝑜𝑛𝑠 𝑎𝑡 𝑧 = 0 𝑏𝑢𝑡 𝑓 ′ 0 𝑑𝑜𝑒𝑠 𝑛𝑜𝑡 𝑒𝑥𝑖𝑠𝑡𝑠.

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