Degree 5 zeros 8,i,6i Enter the remaining zeros
2 answers:
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Answer:
θ = -33.69°
Step-by-step explanation:
For Φ>0 and Φ<0 (in general Φ≠nπ where n is an integer), sin(Φ) ≠ 0
Dividing both equations:
Therefore:
arctan(θ) = -2/3
θ = -33.69°
The answer does not depend on the sign of Φ, in fact we just need that the sine does not become zero, which occurs when Φ is equal to an integer times π (radians) or 180 (degrees)
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Answer:
The required polynomial is .
Step-by-step explanation:
If a polynomial has degree n and are zeroes of the polynomial, then the polynomial is defined as
It is given that the polynomial R has degree 4 and zeros 3 − 3i and 2. The multiplicity of zero 2 is 2.
According to complex conjugate theorem, if a+ib is zero of a polynomial, then its conjugate a-ib is also a zero of that polynomial.
Since 3-3i is zero, therefore 3+3i is also a zero.
Total zeroes of the polynomial are 4, i.e., 3-3i, 3_3i, 2,2. Let a=1, So, the required polynomial is
Therefore the required polynomial is .