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Hi, i was stuck on a question and think i realised how to do it while i was writing it here. But im still not 100% sure i did it right. so....
Z1 = 9+7j
Z2 = 8+3j
find |Z1| - |Z2|
is the answer 19?
did i get the rite answer? Please Help
Lloyd
Putting a value like |this| means that you want the magnitude of it.
It doesn't matter what the direction is, it just wants how big it is.
Real and imaginary numbers are plotted at right-angles to each other, so you can use pythagoras.
|Z1| - |Z2| = |9+7j| - |8+3j| = √(9²+7²) - √(8²-3²)
= √130 - √73 ≈ 2.86
Ugly but true.
Why did the vector cross the road?
It wanted to be normal.
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j? My book always used i!
Btw, is there any practical use for imaginary numbers at all?
A logarithm is just a misspelled algorithm.
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i and j are interchangeable.
Imaginary numbers can be helpful in engineering and engineers use j a lot because I would mean current, so using i would get confusing.
Why did the vector cross the road?
It wanted to be normal.
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I did a unit on electronics at uni. Failed the first test, but scraped through on final exam.
In that course, our prof said that the first person to use imaginary numbers in electonics was laughed at! His colleagues said "Imaginary? but circuits are Real ... hahahaha!"
So he used the maths to calculate a circuit which looked normal, but would in fact short-circuit. He then put the resistors and capacitors together, showed it to his colleagues, and said "Lo! A short-circuit!" They all laughed again.
So he said, "OK, attach a battery then!"
Phhzzztt!
They have used imaginary numbers ever since.
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
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