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Maths Extension 2 Predictions/Thoughts (3 Viewers)

synthesisFR

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wait but like for 7
how is it b? if
theta = pi, and pi/2 then u get arg (theta1theta2) as 3pi/2 which is not principal arg? thats why i put C
 

pdawg

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its b. B does not have a definitive frequency for shm

A is actually periodic in pi
what if theta1 is pi/2 and theta2 is 2pi/3. Arg(z1z2) would be equal to -5pi/6 and theta1 + theta2 = 7pi/6. am i stupid or sum
 

scaryshark09

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Arg(ab) = Arg(a) + Arg(b)
that is the formula you have to use i think
thats how i got b
 

notme123

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tan(theta) = y/x does imply x + iy = rexp(itheta) for only ONE solution of tan(theta) = y/x

its which solution that matters but its irrelevant for this question because its in the 'if' part, not the 'then' part
 

synthesisFR

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Arg(ab) = Arg(a) + Arg(b)
that is the formula you have to use i think
thats how i got b
yeah but like does this counter example not disprove it since the final arg is not principal which is what the question states..
wait but like for 7
how is it b? if
theta = pi, and pi/2 then u get arg (theta1theta2) as 3pi/2 which is not principal arg? thats why i put C
 

notme123

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heres an example,

tan(-pi/4) = 1/-1 which implies that -1+i = r*exp(-i pi/4) but only when r = -sqrt(2)
 

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