HW Solution 7 - lVcV> <V AW...

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Unformatted text preview: lVcV> <V AW " V M-z V v . ^ X-tr C i r * r f o ^ <s > o ^-< \^ r " - '~ J <^->l --v ? yP u; lf- ? t>-\ o " Sk c i-cr x (/> T \4- H r M M C \ O O U O]D u r ^ i \ 2 T .-5 - ^ O T ^I- *\ *r* \ T _ = t ^ C= Q - ^ -V ^ ^K- h ' C D d >i\^ tM, PC,-^_J P VY\ V t :.>vu\ O-V U U 3 42, V M I f ^ H \ UTV J J GA 2.-V- <>o Y\ & %l v> & T- L.OV M ^ fi'A _ __ .- * ^ _3 o *KA , ex Cn "I T 3 i H T '' VvX^\ S4>n^-jp APPE&DIX ALPHA: Determining the n th root of -J Finding the n th root of a negative number requires that we recall how to represent it as a complex number in both polar and exponential form. What I mean is that a complex number given in rectangular form bi a + can be represented in either polar or exponential form via the following: [ ] i re i r bi a = + = + sin cos where 2 2 b a r + = . (1) We can think of J as a complex number using Eq. (1) by nothing that J a = and = b , which upon substitution into Eq. (1) yields: [ ] i Je i J J = + = sin cos . (2) The most obvious way for the first equality to be true is that the polar angle should equal to , leading to [ ] i...
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HW Solution 7 - lVcV> <V AW...

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