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

1. 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?
2. (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.
3. (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 ? That's patently untrue. We have , now do the same for your expression, except with and .
4. I see.
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?
5. (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 ? That's patently untrue. We have , now do the same for your expression, except with and .
Thanks Alexion and Zacken. Zacken is this what you implied as well (what Alexion said)?

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