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how to reduce (4.8+j1.6) / (160-j80) by hand the answer is 20+j20 but I can't seem to work it out
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It's definitely not that. (4.8+j1.6) / (160-j80) = (48+16j)/10 . 1/80 . 1/(2-j) = 1/800 . (48 + 16j) . (2+j)/(2^2 - i^2) = 1/4000 . (48 + 16j)(2+j) = 1/4000 . (96-16 + (48+32).j) = 1/4000 . (80 + 80j) = 0.02 + 0.02i
Yeah you are right they converted it to mili without showing that step. Thanks
last line should be j of course; old habits die hard.
haha no worries it all make sense thanks!
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