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Question: If the real part of the complex number $z = \frac{3 + 2 i \cos θ}{1 − 3 i \cos θ}$, $θ ∈ ( 0 , \fra...

If the real part of the complex number z=3+2icos⁡θ1−3icos⁡θz = \frac{3 + 2 i \cos θ}{1 − 3 i \cos θ}, θ∈(0,π2)θ ∈ ( 0 , \frac{π}{2} ) is zero, then the value of sin⁡23θ+cos⁡2θ\sin^2 3 θ + \cos^2 θ is equal to _____

Answer

1

Explanation

Solution

To find the value of the expression, we first need to determine the value of θ\theta. The given complex number is z=3+2icos⁡θ1−3icos⁡θz = \frac{3 + 2 i \cos θ}{1 − 3 i \cos θ}. To find its real part, we multiply the numerator and the denominator by the conjugate of the denominator, which is 1+3icos⁡θ1 + 3i \cos θ.

z=(3+2icos⁡θ)(1+3icos⁡θ)(1−3icos⁡θ)(1+3icos⁡θ)z = \frac{(3 + 2 i \cos θ)(1 + 3 i \cos θ)}{(1 − 3 i \cos θ)(1 + 3 i \cos θ)}

Expand the numerator: (3)(1)+(3)(3icos⁡θ)+(2icos⁡θ)(1)+(2icos⁡θ)(3icos⁡θ)(3)(1) + (3)(3i \cos θ) + (2i \cos θ)(1) + (2i \cos θ)(3i \cos θ) =3+9icos⁡θ+2icos⁡θ+6i2cos⁡2θ= 3 + 9i \cos θ + 2i \cos θ + 6i^2 \cos^2 θ Since i2=−1i^2 = -1, this becomes: =3+11icos⁡θ−6cos⁡2θ= 3 + 11i \cos θ - 6 \cos^2 θ =(3−6cos⁡2θ)+11icos⁡θ= (3 - 6 \cos^2 θ) + 11i \cos θ

Expand the denominator using the formula (a−b)(a+b)=a2−b2(a-b)(a+b) = a^2 - b^2: 12−(3icos⁡θ)21^2 - (3i \cos θ)^2 =1−9i2cos⁡2θ= 1 - 9i^2 \cos^2 θ Since i2=−1i^2 = -1, this becomes: =1−9(−1)cos⁡2θ= 1 - 9(-1) \cos^2 θ =1+9cos⁡2θ= 1 + 9 \cos^2 θ

So, the complex number zz is: z=(3−6cos⁡2θ)+11icos⁡θ1+9cos⁡2θz = \frac{(3 - 6 \cos^2 θ) + 11i \cos θ}{1 + 9 \cos^2 θ}

Now, we separate the real and imaginary parts of zz: Re(z)=3−6cos⁡2θ1+9cos⁡2θ\text{Re}(z) = \frac{3 - 6 \cos^2 θ}{1 + 9 \cos^2 θ} Im(z)=11cos⁡θ1+9cos⁡2θ\text{Im}(z) = \frac{11 \cos θ}{1 + 9 \cos^2 θ}

The problem states that the real part of zz is zero: Re(z)=0\text{Re}(z) = 0 3−6cos⁡2θ1+9cos⁡2θ=0\frac{3 - 6 \cos^2 θ}{1 + 9 \cos^2 θ} = 0

Since the denominator 1+9cos⁡2θ1 + 9 \cos^2 θ is always positive (as cos⁡2θ≥0\cos^2 θ \ge 0, so 1+9cos⁡2θ≥11 + 9 \cos^2 θ \ge 1), the numerator must be zero: 3−6cos⁡2θ=03 - 6 \cos^2 θ = 0 3=6cos⁡2θ3 = 6 \cos^2 θ cos⁡2θ=36\cos^2 θ = \frac{3}{6} cos⁡2θ=12\cos^2 θ = \frac{1}{2}

We are given that θ∈(0,π2)\theta \in (0, \frac{\pi}{2}). In this interval, cos⁡θ\cos θ is positive. So, cos⁡θ=12=12\cos θ = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}. For cos⁡θ=12\cos θ = \frac{1}{\sqrt{2}} in the interval (0,π2)(0, \frac{\pi}{2}), the value of θ\theta is π4\frac{\pi}{4}.

Now we need to find the value of sin⁡23θ+cos⁡2θ\sin^2 3\theta + \cos^2 \theta. Substitute θ=π4\theta = \frac{\pi}{4} into the expression: sin⁡2(3⋅π4)+cos⁡2(π4)\sin^2 \left(3 \cdot \frac{\pi}{4}\right) + \cos^2 \left(\frac{\pi}{4}\right) =sin⁡2(3π4)+cos⁡2(π4)= \sin^2 \left(\frac{3\pi}{4}\right) + \cos^2 \left(\frac{\pi}{4}\right)

We know that sin⁡(3π4)=sin⁡(π−π4)=sin⁡(π4)=12\sin\left(\frac{3\pi}{4}\right) = \sin\left(\pi - \frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}. And cos⁡(π4)=12\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}.

Substitute these values into the expression: (12)2+(12)2\left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 =12+12= \frac{1}{2} + \frac{1}{2} =1= 1