Precalculus - DeMoivre's Theorem
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@vivianka216 You've probably been told the answer for this already but in case you haven't: xD
(1+1i)^16... so the modulus (or 'r') is root(1+1) or root(2).
The argument, or the angle 1+i makes with the real axis on an argand diagram, would be 45 degrees (because both 1 and 1i are the same length) and 45 degrees is the same as pi/4 radians... so yeah then he just subs them into r(cos(theta)+isin(theta))
root(2)(cos(pi/4)+isin(pi/4))
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Id be seriously distracted with a maths teacher looking like that.......very clear explanation though!
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his muscles are distracting me. I can x-ray thru his shirt. >.<
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pi/4 comes from a triangle with side=1 unit and height=1 unit. Thus the angle formed is 45 degree or pi/4. (1 unit at the x-axis and 1 unit at the i-axis)
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He should probably explain how he got the argument pi/4 instead of pulling it out of thin air. I can see that tripping up a lot of people.
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@vivianka216 He came to that conclusion based upon two different formulas that are used to convert cartesian coordinates (x,y)/(a,b) into polar coordinates (r,angle). To put it simply, the r in trigonometric form equals the square root of a^2 + b^2 from the rectangular form (a + bi). The angle in trigonometric form equals the inverse tangent of b/a from the rectangular form.
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@vivianka216 that's (1+i)^16, if u look at the point 1+i on an argand diagram, r = (1^2+1^2)^(1/2) which means the modulus is the square root of 2. The argument is therefore pi/4 bcos tan 1/1 = 45
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@vivianka216 that's (1+i)^16, if u look at the point 1+i on an argand diagram, r = (1^2+1^2)^(1/2) which means the modulus is the square root of 2. The argument is therefore pi/4 bcos tan 1/1 = 45
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You lost me at (b)
lovely2626 1 year ago 7
for problem b, how did you find the trig form of (1+6)^16?
vivianka216 1 year ago 5