Write 3 - (i^3 - i + 2)/(i^2 - i^4) in the form x + yi, where x and y are real numbers. Please show any working out thanks!
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Originally Posted by Joker37 Write 3 - (i^3 - i + 2)/(i^2 - i^4) in the form x + yi, where x and y are real numbers. Please show any working out thanks! $\displaystyle 3-\frac{i^3-i+2}{i^2-i^4}$ (Did I interpret this correctly??) Note that $\displaystyle i^2=-1,~i^3=-i,~i^4=1$ Can you try to continue on?
Originally Posted by Chris L T521 $\displaystyle 3-\frac{i^3-i+2}{i^2-i^4}$ (Did I interpret this correctly??) Note that $\displaystyle i^2=-1,~i^3=-i,~i^4=1$ Can you try to continue on? 3 - (i^3 - i + 2)/(i^2 - i^4) 3 - (-i -1 +2)/(-1 -1) 3 - (-i + 1)/-2 (3/1) - (-i + 1)/-2 (6/2) - (i + 1)/2 Is this right so far?
Originally Posted by Joker37 3 - (i^3 - i + 2)/(i^2 - i^4) 3 - (-i -1 +2)/(-1 -1) Mr F says: This should be 3 - (-i - i +2)/(-1 -1). You will need to fix what follows. 3 - (-i + 1)/-2 (3/1) - (-i + 1)/-2 (6/2) - (i + 1)/2 Is this right so far? ..
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