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1/3-i-1/1+2i
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\[\frac{ 1 }{ 3-i }-\frac{ 1 }{ 1+2i }\]
(1/10) + (1/2)i
how did you get that
@randomguyinternet plz respond
is i an imaginary number?
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yeah
i got an asnwer but i just wunna check it..i got a diff one that the other guy
so okay, it's easy
1(1+2i) - 1(3-i)
simplify that part first
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(1 / 3) - i - (1 / 1) + (2 * i) = -0.666666667 + i
it shud be -2+3i
now simplify the denominator of, (3-i)(1+2i)
\[\frac{ 1+5i }{ 10 }\] is what i got :/
no, it shud be \[\frac{ -2+3i }{ 5+5i }\]
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hmmm well you can't have a radical on bottom...and i is a radical number
ok ok i got chu
so multiply by 5-5i, both top and bottom
(1+5i)/10 is correct
so i was right lol
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