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

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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Avatar for Alexion
Alexion
17
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.
(edited 8 years ago)
Reply 2
Avatar for Zacken
Zacken
22
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∣=a2+i2b2|a+ib| = \sqrt{a^2 + i^2b^2}? That's patently untrue. We have ∣a+ib∣=a2+b2|a+ib| = \sqrt{a^2 + b^2}, now do the same for your expression, except with a=cos⁡θa = \cos \theta and b=sin⁡θb = \sin \theta.
I see. ∣a+ib∣=a2+b2|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?
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∣=a2+i2b2|a+ib| = \sqrt{a^2 + i^2b^2}? That's patently untrue. We have ∣a+ib∣=a2+b2|a+ib| = \sqrt{a^2 + b^2}, now do the same for your expression, except with a=cos⁡θa = \cos \theta and b=sin⁡θb = \sin \theta.


Thanks Alexion and Zacken. Zacken is this what you implied as well (what Alexion said)?

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