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sin (z) = 2, Is There Any Root Possible For z ?

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  • Difficulty Level : Expert
  • 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 :

e = 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 = re, where
z = a + ib, 
r = |a2 + b2| 
θ = tan-1(b/a),
So, taking log on both sides of the equation z = re  
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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