ee20-hw01-s08-sol - EECS 20N Structure and Interpretation...

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EECS20N:StructureandInterpretationofSignalsandSystems Department of Electrical Engineering and Computer Sciences U NIVERSITY OF C ALIFORNIA B ERKELEY Problem Set 1 SOLUTIONS HW1.1 (a) z 1 = 1 + i 3 = 2( 1 2 + i 3 2 ) = 2exp( i π 3 ) z 2 = exp( i 2 π 3 ) = cos( 2 π 3 ) + i sin( 2 π 3 ) = 1 2 + i 3 2 See Fig. 1 2 1 1/2 z1 z1* 1/z1* 1/z1 z2 = 1/z2* z2* = 1/z2 Real Axis Imaginary Axis -1/2 -1 -2 Figure 1: Problem 1.1a (b) z 1 + 2 z 2 = 1 1 + i ( 3 + 3) = i 2 3 z 2 1 + z 2 = (2exp( i π 3 )) 2 + exp( i 2 π 3 ) = 4exp( i 2 π 3 ) + exp( i 2 π 3 ) = 5exp( i 2 π 3 ) 1 2 z 1 + z * 2 = 1 2 (1 + i 3) + ( 1 2 + i 3 2 ) * = 1 2 + i 3 2 1 2 i 3 2 = 0 (c) | z 1 | = | 2exp( i π 3 ) | = 2 | z 2 | = | exp( i 2 π 3 ) | = 1 | z 1 z 2 | = | z 1 || z 2 | = 2 | z 1 z * 2 | = | z 1 || z * 2 | = | z 1 || z 2 | = 2 1
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| z 1 /z 2 | = | z 1 | / | z 2 | = 2 | z 2 /z 1 | = | z 2 || z 1 | = 1 / 2 (d) (i) z 2 1 = (2exp( i π 3 )) 2 = 4exp( i 2 π 3 ) (ii) z 3 1 = (2exp( i π 3 )) 3 = 8exp( i 3 π 3 ) = 8 (iii) z 6 1 = (2exp( i π 3 )) 6 = 64exp( i 6 π 3 ) = 64exp( i 2 π ) = 64 (iv) z 4 2 = (exp( i 2 π 3 )) 4 = exp( i 8 π 3 ) = exp( i (2 π + 2 π 3 )) = exp( i 2 π 3 ) = z 2 (e) z 1 / 4 2 = exp( i (2 + 2 π 3 )) 1 / 4 n ∈ { 0 , 1 , 2 , 3 } z 1 / 4 2 = exp( i ( 2 π 3 )) 1 / 4 , exp( i (2 π + 2 π 3 )) 1 / 4 , exp( i (4 π + 2 π 3 )) 1 / 4 , exp( i (6 π + 2 π 3 )) 1 / 4 z 1 / 4 2 = exp( i π 6 ) , exp( i ( π/ 2 + π 6 )) , exp( i ( π + π 6 )) , exp( i (3 π/ 2 + π 6 )) z 1 / 4 2 = exp( i π 6 ) , exp( i ( 2 π 3 )) , exp( i ( 7 π 6 )) , exp( i ( 10 π 6 )) See Figure 2. The circles are the roots and the cross is z 2 . 1 Real Axis Imaginary Axis -1 0 Figure 2: Problem 1.1e 2
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HW1.2 1. Using the given hint we write e i 2 θ = ( e ) 2 , so cos(2 θ ) + i sin(2 θ ) = cos( θ ) 2 sin( θ ) 2 + i 2cos( θ )sin( θ ) The result follows by comparing the real and imaginary parts of this equality: cos(2 θ ) = cos( θ ) 2 sin( θ ) 2 and sin(2 θ ) = 2cos( θ )sin( θ ) .
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