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can someone show how modulus of z = r (fp1 Complex numbers) watch

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    If z = r(cos a + isin a)
    Then modulus of z = r x sqareroot of(cos^2 a + i^2 sin^2 a)
    But I don't see how sqare root of the the above thing will be 1. What should the step actually look like?
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    (Original post by thebrahmabull)
    If z = r(cos a + isin a)
    Then modulus of z = r x sqareroot of(cos^2 a + i^2 sin^2 a)
    But I don't see how sqare root of the the above thing will be 1. What should the step actually look like?
    It's a standard trigonometric identity that cos2x + sin2x = 1. Derived from Pythagorean principles.

    Remember that the i2 isn't included as sina is the length of the imaginary vector of z.
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    (Original post by thebrahmabull)
    If z = r(cos a + isin a)
    Then modulus of z = r x sqareroot of(cos^2 a + i^2 sin^2 a)
    But I don't see how sqare root of the the above thing will be 1. What should the step actually look like?
    Do you think that |a+ib| = \sqrt{a^2 + i^2b^2}? That's patently untrue. We have |a+ib| = \sqrt{a^2 + b^2}, now do the same for your expression, except with a = \cos \theta and b = \sin \theta.
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    I see. |a+ib| = \sqrt{a^2 + b^2}
    But what happened to the i up there? OK just remembered : the modulus function converts everything to it's magnitude . so the minus due to i^2 got converted to + :is that what happened?
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    (Original post by Alexion)
    It's a standard trigonometric identity that cos2x + sin2x = 1. Derived from Pythagorean principles.

    Remember that the i2 isn't included as sina is the length of the imaginary vector of z.
    (Original post by Zacken)
    Do you think that |a+ib| = \sqrt{a^2 + i^2b^2}? That's patently untrue. We have |a+ib| = \sqrt{a^2 + b^2}, now do the same for your expression, except with a = \cos \theta and b = \sin \theta.
    Thanks Alexion and Zacken. Zacken is this what you implied as well (what Alexion said)?
 
 
 
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