Sometimes it is real
If z + z 1 = cos i + i sin i , then
z 3 + z 3 1 = e 3 A − B e 2
where A and B are positive integers, what is A + B ?
Notations:
- i = − 1 denotes the imaginary unit .
- e ≈ 2 . 7 1 8 denotes the Euler's number .
The answer is 4.
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1 solution
But is i in radians?
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i = − 1 is the imaginary unit.
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Ya i know. But here, the euler's theorem
cos θ + i sin θ = e i θ where θ is in radians, isnt it?
how is theta = i applicable? We can say that 1 8 0 i π in radians.
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@Md Zuhair – Ah, I see what you're saying. θ is a variable used often because of the relation of Euler's formula to trigonometric functions, but the general theorem states that for any complex number x ,
e i x = cos x + i sin x
Since x = i is complex, the theorem still holds, so Chew's solution remains.
@Md Zuhair – Yes, it is in radians. But since the radian is the ratio of two lenghts, it's not really a unit, it's a pure number, so θ can be any number, even i
Relevant wiki: Euler's Formula
z + z 1 ( z + z 1 ) 3 z 3 + 3 z + z 3 + z 3 1 ( z + z 1 ) 3 z 3 + z 3 1 = cos i + i sin i = e i 2 = e − 1 = e 1 = e 3 1 = e z 1 = e 3 1 = e 3 1 − 3 ( z + z 1 ) = e 3 1 − e 3 = e 3 1 − 3 e 2 By Euler’s formula: e i θ = cos θ + i sin θ
⟹ A + B = 1 + 3 = 4