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Chem Differential Eq HW Solutions Fall 2011 155

# Chem Differential Eq HW Solutions Fall 2011 155 - z and Log...

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Section 12.5 Analytic Functions 155 Solutions to Exercises 12.5 1. (a) (3 + 2 i )(2 - i ) = 6 - 3 i + 4 i - 2 i 2 = 8 + i . (b) (3 - i ) (2 - i ) = (3 - i )(2 + i ) = 7 + i . (c) 1 + i 1 - i = 1 + i 1 - i · 1 - i 1 - i = (1 + i )(1 + i ) 1 2 + 1 2 = i. 5. (a) Arg ( i ) = π 2 . (b) | i | = 1. (c) i = e i π 2 . 9. (a) Arg (1 + i ) = π 4 . (b) | 1 + i | = 1 2 + 1 2 = 2. (c) 1 + i = 2 e i · π 4 . In computing the values of Arg z , just remember that Arg z takes its values in the interval ( - π,π ]. Consequently, Arg z is not al- ways equal to tan - 1 ( y/x ) (see Section 12.5, (8), for the formula that relates Arg z to tan - 1 ( y/x )). You can use Mathematica to evaluate Arg z and the absolute value of z . This is illustrated by the following exercises. 17. (a) Apply Euler’s identity, e 2 i = cos 2 + i sin2. (b) Use Example 1(d): for z = x + iy , sin z = sin x cosh y + i cos x sinh y. So sin i = sin 0 cosh 1 + i cos 0 sinh 1 = i sinh 1 = i e - e - 1 2 . (c) Use Example 1(e): for z = x + iy , cos z = cos x cosh y - i sin x sinh y. So cos i = cos 0 cosh 1 - i sin 0 sinh1 = cosh 1 e + e - 1 2 . (It is real!) (d) Use Example 1(f): for z = x + iy , Log z = ln( | z | ) + i Arg z. For z = i , | i | = 1 and Arg i = π 2 . So Log i = ln 1 + i π 2 = i π 2 , because ln 1 = 0.
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Unformatted text preview: z , and Log z is one of them. All other values of log z diﬀer from Log z by an integer multiple of 2 πi . This is because the imaginary part of the logarithm is de±ned by using a branch of arg z , and the branches of arg z diﬀer by integer multiples of 2 π . (See Applied Complex Analysis and PDE for more details on the logarithm.) In particular, the imaginary part of Log z , which is Arg z , is in the interval (-π, π ]. You can use Mathematica to evaluate Log z and e z . This is illus-trated by the following exercises. 21. We have (-1) · (-1) = 1 but 0 = Log 1 ± = Log (-1) + Log (-1) = iπ + iπ = 2 iπ. 25. (a) By de±nition of the cosine, we have cos( ix ) = e i ( ix ) + e-i ( ix ) 2 = e-x + e x 2 = cosh x....
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