sin (z) = 2, Is There Any Root Possible For z ?
Last Updated :
08 Feb, 2021
A question arises that is there any root possible for z if sin (z) = 2. First answer which comes to mind that no such roots are possible, as
-1 ≤ sin θ ≤ 1
But if we go in deep, we’ll find that although no real roots are possible for z, only imaginary (or complex) roots are possible.
So, let’s find the solution :
According to Euler’s form :
eiθ = cosθ + i*sinθ
where, e = base of the natural logarithm
i = imaginary part , ( i = √(-1) )
θ = the angle in radian
So in the above formula, substitute ∅ = z
So, the equation becomes
eiz = cos z + i sin z —–( i )
Now, put z = -z
ei(-z) = cos(-z) + i sin(-z)
e-iz = cos z – i sin z —–( ii ) [ cos(-θ) = cosθ , sin(-θ) = -sinθ ]
Now, subtracting equation ( i ) by ( ii )
eiz – e-iz = (cos z + i sin z) – (cos z – i sin z)
eiz – e-iz = 2i sin z
eiz – e-iz = 2i *(2) [ Since, sin z = 2 (given) ]
eiz – 1/eiz = 4i
Multiplying both side of the equation by eiz
e2(iz) – 1 = 4i eiz
e2(iz) – 4i eiz – 1 = 0
Now, let y = eiz
y2 – 4iy – 1 = 0
Now, to find the roots of y in the above quadratic equation, we apply Dharacharya formula, which says :
A quadratic equation of the form ax2 + bx + c = 0, a ≠ 0
then roots of x are, x = (-b ± √(b2-4ac)/2a
So, applying Dharacharya formula in the above equation,
y = ( 4i ± √(16i2 + 4 ) ) / 2
y = ( 4i ± √(-16 + 4 ) ) / 2 [i2 = -1]
y = ( 4i ± √(-12) ) / 2
y = ( 4i ± 2√3i ) / 2 [√(-12) = √(-1*12) = √(-1)*√12) = i * 2√3, Since √(-1) = i ]
y = 2i ± √3i
Now, putting back the value of y = eiz in the above equation
eiz = 2i ± √3i
eiz = i * (2 ± √3)
Now, taking log(ln) on both sides
iz = ln(i) + ln(2 ± √3) —–( iii ) [ln(ea) = a, and ln(a*b) = ln(a) + ln(b)]
Now for solving ln(i), we have to understand the following concept :
In polar representation of complex numbers, we write z = reiθ, where
z = a + ib,
r = |a2 + b2|
θ = tan-1(b/a),
So, taking log on both sides of the equation z = reiθ
ln(z) = ln(r) + iθ [ln(ea) = a, and ln(a*b) = ln(a) + ln(b)]
Putting the value of z, r and θ in the above equation
ln(a+ib) = ln(|a2 + b2|) + i*tan-1(b/a)
So, writing ln(i) = ln(0 + 1i), and applying the above formula
ln(0+1i) = ln( |02 + 12| ) + i*tan-1(1/0)
ln(i) = ln1 + i*∏/2 [ tan-1(1/0) = tan-1(∞) = ∏/2 ]
ln(i) = i*∏/2 [ ln1 = 0 ]
Now putting the value of ln(i) in equation ( iii )
iz = i*∏/2 + ln(2 ± √3)
Dividing both sides of the equation by i
z = ∏/2 + ln(2 ± √3)/i
z = ∏/2 + ( ln(2 ± √3) * i )/ ( i * i) [Dividing numerator and denominator by i]
z = ∏/2 – i * ln(2 ± √3) [i2 = -1]
z = ∏/2 – i * ln(2 ± √3)
So, we got one complex root, but more are there,
as sin (θ ± 2n∏) = sin θ , n = 1,2,3,..
So,
z = ∏/2 – i * ln(2 ± √3) ± 2n∏ , n = 1,2,3,….
Therefore, now we get that for sin(z) = 2, there are infinite complex roots of z.
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