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hw02fa13-sol

# hw02fa13-sol - ECE 2026 Fall 2013 Solutions to Problem...

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ECE 2026 Fall 2013 Solutions to Problem Set #2 1) Convert the following to polar form: a. 𝑧𝑧 = 𝑒𝑒 ± 𝑒𝑒 ±² / ± 𝑒𝑒 ± ±² / ± = 𝑒𝑒 ± 𝑗𝑗 ( 𝑗𝑗 ) = 𝑗𝑗 2 𝑒𝑒 ± = 2 𝑒𝑒 ± 𝑒𝑒 ±² / ± where in several places we have used the equalities 𝑗𝑗 = 𝑒𝑒 ±² / ± and 𝑗𝑗 = 𝑒𝑒 ± ± ± / ± . b. 𝑧𝑧 = 1 + 𝑒𝑒 ± ± ± . Pull a factor of 𝑒𝑒 ± ± ± out of both terms: 𝑧𝑧 = 𝑒𝑒 ± ± ± ± + 𝑒𝑒 ± ± ± 𝑒𝑒 ± ± ± . Then use Euler’s identity to spot 2 times a cosine inside the parens: 𝑧𝑧 = 2 cos ( 4 𝜔𝜔 ) 𝑒𝑒 ± ± ± . c. Direct approach : Look at the sum and notice that it has three pairs that are negatives of each other. For example, 𝑒𝑒 ± ± ± / ± = 𝑒𝑒 ± ( ± ± ± ± / ± ) = 𝑒𝑒 ±² 𝑒𝑒 ± ± ± / ± = 𝑒𝑒 ± ± ± / ± . Each of these pairs thus sums to zero and the only term left is 𝑒𝑒 ± ± ± / ± = 𝑒𝑒 ±² / ± . Approach using the n’th roots of unity : Note that this is a sum of powers of 𝑒𝑒 ±² / ± , which is one of the 8 ± ± roots of unity. The geometric series has a closed form given by 𝑧𝑧 ± ± ± ± ± ± ± = ( 𝑧𝑧 ± 1 ) ( 𝑧𝑧 1 ) . When you put 𝑧𝑧 = 𝑒𝑒 ±² / ± in this sum, you get zero (verify this!). Note that one term, 𝑒𝑒 ± ± ± / ± , is missing from the series given in this problem. Thus, this series sums to 0 𝑒𝑒 ± ± ± / ± = 𝑒𝑒 ±² 𝑒𝑒 ±² / ± = 𝑒𝑒 ±² / ± .

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