+0

# ln ( 10.75 + 6.98i )

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ln ( 10.75 + 6.98i )

Guest Jan 15, 2015

#1
+18829
+10

ln ( 10.75 + 6.98i )

$$\small{\text{ z=10.75+6.98\ i \quad \ln{(z)} = ? }}\\ \small{ Solution: \boxed{ln{(z)} = \ln{(|z|)} + i*(arg(z)+2\pi k ) } } \\\\ \small{\text{ I.  |z| = \sqrt{10.75^2+6.98^2} = 12.8172891050  }}\\ \small{\text{ II.  \phi\ensurement{^{\circ}} = \tan^{-1}{(\frac{6.98}{10.75} )} = 32.9957575073\ensurement{^{\circ}}  }}\\ \small{\text{ III.  arg(z) = \phi \ensurement{^{\circ}} * \frac{ \pi }{ 180\ensurement{^{\circ}} }+2\pi k = 0.57588460769 + 2\pi k }}\\\\ \small{\text{  \ln{(z)} = \ln{(12.8172891050 )}+i*(0.57588460769+2\pi k)  }}\\ \small{\text{  \boxed{ \ln{(10.75+6.08\ i)} = 2.55079497086+(0.57588460769+2\pi k) \ i } \quad k=0,1,2\dots  }}$$

heureka  Jan 15, 2015
Sort:

#1
+18829
+10

ln ( 10.75 + 6.98i )

$$\small{\text{ z=10.75+6.98\ i \quad \ln{(z)} = ? }}\\ \small{ Solution: \boxed{ln{(z)} = \ln{(|z|)} + i*(arg(z)+2\pi k ) } } \\\\ \small{\text{ I.  |z| = \sqrt{10.75^2+6.98^2} = 12.8172891050  }}\\ \small{\text{ II.  \phi\ensurement{^{\circ}} = \tan^{-1}{(\frac{6.98}{10.75} )} = 32.9957575073\ensurement{^{\circ}}  }}\\ \small{\text{ III.  arg(z) = \phi \ensurement{^{\circ}} * \frac{ \pi }{ 180\ensurement{^{\circ}} }+2\pi k = 0.57588460769 + 2\pi k }}\\\\ \small{\text{  \ln{(z)} = \ln{(12.8172891050 )}+i*(0.57588460769+2\pi k)  }}\\ \small{\text{  \boxed{ \ln{(10.75+6.08\ i)} = 2.55079497086+(0.57588460769+2\pi k) \ i } \quad k=0,1,2\dots  }}$$

heureka  Jan 15, 2015
#2
+91458
0

Thanks Heureka, that looks impressive

Melody  Jan 15, 2015
#3
+26402
+5

Here's an alternative approach (I've just considered the principal solution):

.

Alan  Jan 16, 2015

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