Solving A Very Radical Equation | Problem 337

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[XJWill]

Here is the proper way to solve this equation. Begin by making the sensible assumption that every function in the equation is single-valued, using the principal value where necessary. Next let
z = r * exp(i * t) where real-valued r > 0 and t is real-valued and - pi < t <= pi
Since the principal value of the complex square-root is in the 1st or 4th quadrant of the complex plane, this means that
sqrt(z) = sqrt(r) * exp(i * t / 2)
but the other square-root is more complicated. Since
- z = - r * exp(i * t)
= r * exp( i * (t + pi) )
then the square-root is
sqrt( - z ) = sqrt(r) * exp(i * (t + pi) / 2 ) * s where s is either +1 or -1 depending on t
= sqrt(r) * exp(i * t / 2) * exp(i * pi / 2) * s
= sqrt(r) * exp(i * t / 2) * i * s
Note that if t is in the 1st or 2nd quadrant of the complex plane then
s = - 1 t in 1st or 2nd quadrant
because t/2 + pi/2 is in the 2nd quadrant and it needs to be shifted to the 4th quadrant.
But if t is in the 3rd or 4th quadrant then
s = + 1 t in 3rd or 4th quadrant
because t/2 + pi/2 is in the 1st quadrant and does not need to be shifted.
Finally, if t = 0 then s = 1 and if t = pi then s = - 1
These preliminaries take some time to write out explicitly, but once you understand them, you do not usually need to write them out. Now to solve the equation:
sqrt(z) - sqrt( - z ) = 6 + i*8
sqrt(r) * exp(i * t / 2) - sqrt(r) * exp(i * t / 2) * i * s = 6 + i*8
sqrt(r) * exp(i * t / 2) * (1 - i*s) = 6 + i*8
Comparing the square of the modulus of both sides gives
r * 2 = 36 + 64
r = 50
Comparing the argument of both sides requires care to account for the value of s. Separate into two cases,
1: s = 1 for - pi < t <= 0
2: s = -1 for 0 < t <= pi

1: sqrt(r) * exp(i * t / 2) * (1 - i) = 6 + i*8
This equation can be seen to have no solution with t in the assumed range since t/2 is 4th quadrant
and (1 - i) is 4th quadrant while 6 + i*8 is 1st quadrant.
2: sqrt(r) * exp(i * t / 2) * (1 + i) = 6 + i*8
exp(i * t/2) = 1/sqrt(50) * (6 + i*8) / (1 + i)
exp(i * t/2) = sqrt(2) / 10 * (7 + i)
exp(i * t) = 1/25 * (24 + i*7)
r * exp(i * t) = 50 * exp(i * t)
= 48 + i * 14

[Nobodyman]

You are the best solver complex problems in the world❤✨🌟💫⭐

[NadiehFan]

There are tons of videos on YouTube which discuss and solve equations very similar to this, but the solution methods presented are almost always _incorrect_ even though the correct solutions are found. The reason for this is that the rule √a·√b = √(a·b) which is valid for nonnegative real numbers a and b is tacitly assumed to be valid for complex numbers as well, but this generally _not_ true.

For any nonzero complex number z there are two different complex numbers w₁ and w₂ such that w₁² = w₂² = z, and for this reason √z is often claimed to have two different values w₁ and w₂ for any nonzero complex z.

However, this contradicts the definition of √x for any real nonnegative x as the unique nonnegative real number y such that y² = x. After all, the real numbers are a _subset_ of the complex numbers, so when x = 4 is regarded as a real number, then √x = √4 has the unique value 2, but when z = 4 is regarded as a complex number, then √z = √4 would all of a sudden have two values, 2 and −2.

To avoid this inconsistency, √z needs to be defined in such a way that it represents a unique value for any complex number z which also coincides with the usual nonnegative real value in case z happens to be a nonnegative real number. This is exactly what the definition of the _principle_ square root of a complex number entails.

For any complex number z there exists a unique r ∈ [0, ∞) as well as a φ ∈ (−π, π] such that z = r·exp(φi) where φ is unique for any nonzero z. This is of course the familiar exponential form or Euler form of a complex number.

Since the complex number exp(φi) = cos φ + i·sin φ is represented in the complex plane by the point on the unit circle with coordinates (cos φ, sin φ), it follows that z = r·exp(φi) is represented by the point (r·cos φ, r·sin φ) in the complex plane which lies on a circle centered at the origin with radius r and therefore has a distance r from the origin. The distance from the origin of a point in the complex plane representing a complex number z is its _absolute value_ or _modulus_ so we have |z| = r.

The unique value of φ ∈ (−π, π] for any nonzero complex number z = r·exp(φi) = r(cos φ + i·sin φ) = r·cos φ + i·r·sin φ represented by the point (r·cos φ, r·sin φ) in the complex plane is known as the _principal argument_ of the nonzero complex number z.

If we have a half line in the complex plane which starts at the origin and which extends along the positive real axis, then this half line sweeps the entire complex plane if we rotate it around the origin over either half a turn clockwise or half a turn anticlockwise. So, any point in the complex plane can be considered as having the coordinates (r·cos φ, r·sin φ) for a unique r ∈ [0, ∞) as well as a φ ∈ (−π, π] which is unique for any point other than the origin.

Of course, the value φ = −π is excluded because otherwise we would have both φ = π and φ = −π for points on the negative real axis and the value of φ would no longer be unique for those points. If z = 0 then we have |z| = r = 0 and consequently we will then have z = r·exp(φi) for _any_ real value of φ including any φ ∈ (−π, π] meaning that the argument of zero and, a fortiori, the principle argument of zero, is undefined.

The _principal value_ of the square root √z of a complex number z = r·exp(φi) with r ≥ 0, −π < φ ≤ π is _defined_ as

√z = √r·exp(½φi)

where √r is the nonnegative real root of the nonnegative real number r. By convention, for any complex number z (which includes real numbers), the notation √z is commonly (but not universally) understood to refer to this uniquely defined principal value.

We can easily verify that the square of √z = √r·exp(½φi) is (√r·exp(½φi))² = (√r)²·(exp(½φi))² = r·exp(φi) = z as it should be and we can also see that if z is a _positive real number_ then we will have φ = 0 and r = z and consequently we will then have √z = √r·exp(0) = √r so √z will then have its usual positive real value. And, of course, for z = 0 we will have r = 0 and therefore √z = 0.

So, the definition of the principle value of the square root of a complex number satisfies the requirement that it defines a unique value for the square root of any complex number z as well as the requirement that this value coincides with the usual nonnegative real value in case z happens to be a nonnegative real number.

Since the definition of the principal square root of a nonzero complex number z implies that the principal argument of √z is _half_ the principle argument of z it follows that, for _any_ nonzero complex number z, the point representing √z either lies in the right half of the complex plane or on the positive imaginary axis, since −π < φ ≤ π implies −½π < ½φ ≤ ½π. This means that, for any complex z, the real part of √z is always nonnegative.

WolframAlpha also follows the convention that √z represents the principle square root of z for any complex number z (which includes any real number). For example, if you enter √(3 + 4i) or sqrt(3 + 4i) or (3 + 4i)^(1/2) then the (unique) answer is 2 + i, _not_ −2 − i. This also applies to equations with square roots where the radicands are expressions containing variables which can either have real or complex values.

As noted the identity √a·√b = √(a·b) which is valid for nonnegative real numbers a and b is generally _not_ valid for complex numbers and this remains so even with the definition of the square root √z of a complex number z = r·exp(φi) with r ≥ 0, −π < φ ≤ π as √z = √r·exp(½φi).

According to the definition we have √(−1) = i and we also know that for any complex nonreal z the square root √z is represented by a point in the _right half_ of the complex plane. Now, multiplication by √(−1) = i corresponds to a _counterclockwise rotation over a right angle_ around the origin, but this means that the point representing the product √(−1)·√z _may or may not_ lie in the right half of the complex plane. Consequently, it is _not_ guaranteed that √(−1)·√z is equal to √(−z).

If z is a complex number with a _negative_ imaginary part, then the principal argument of z will lie on the interval (−π, 0) and consequently the principal argument of √z will then lie on the interval (−½π, 0) and so the principal argument of the product i·√z = √(−1)·√z will lie on the interval (0, ½π) meaning that √(−1)·√z is a complex number represented by a point in the _upper right half_ of the complex plane.

However, if z is a complex number with a _positive_ imaginary part, then the principal argument of z will lie on the interval (0, π) and consequently the principal argument of √z will then lie on the interval (0, ½π) and so the principal argument of the product i·√z = √(−1)·√z will lie on the interval (½π, π) meaning that √(−1)·√z is a complex number represented by a point in the _upper left half_ of the complex plane. But we know that the square root of any nonreal complex number is represented by a point in the _right half_ of the complex plane, so we can already conclude that √(−1)·√z ≠ √(−z) if Im(z) > 0.

To get to the bottom of this, let's consider a complex nonreal number z = r·exp(φi) with 0 < φ < π so Im(z) > 0, then by definition √z = √r·exp(½φi) with 0 < ½φ < ½π and so √(−1)·√z = i·√z = exp(½πi)·√r·exp(½φi) = √r·exp(½(φ + π)i). On the other hand, if z = r·exp(φi) with 0 < φ < π then we can write −z = r·exp((φ − π)i) with −π < φ − π < 0, so again by definition √(−z) = √r·exp(½(φ − π)i) = −√r·exp(½(φ + π)i) so we have √(−z) = −i·√z = −√(−1)·√z if Im(z) > 0.

Conversely, if z = r·exp(φi) with −π < φ < 0 so Im(z) < 0, then by definition √z = √r·exp(½φi) with −½π < φ < 0 and so √(−1)·√z = i·√z = exp(½πi)·√r·exp(½φi) = √r·exp(½(φ + π)i). But if z = r·exp(φi) with −π < φ < 0 then we can write −z = r·exp((φ + π)i) with 0 < φ + π < π so √(−z) = √r·exp(½(φ + π)i) and we have √(−z) = i·√z = √(−1)·√z if Im(z) < 0.

Since the identity √a·√b = √(a·b) which is valid for nonnegative real numbers a and b is generally _not_ valid for complex numbers we cannot use this to solve an equation such as

√z + √(−z) = 6 + 8i

One approach to avoid invalid manipulations with square roots of complex numbers is to let

√z = w₁, √(−z) = w₂

so

w₁ − w₂ = 6 + 8i

Squaring both sides gives

w₁² + w₂² − 2w₁w₂ = −28 + 96i

and since w₁² = z, w₂² = −z and therefore w₁² + w₂² = 0 this gives 2w₁w₂ = 28 − 96i and so

w₁w₂ = 14 − 48i

From w₁ − w₂ = 6 + 8i and w₁w₂ = 14 − 48i it follows that we have w₁ = −1 + 7i, w₂ = −7 − i or w₁ = 7 + i, w₂ = 1 − 7i. From the first solution set we would get z = w₁² = −48 − 14i but this cannot be a solution of the original equation √z − √(−z) = 6 + 8i if the square roots refer to their principal value (as they should) because the real part of the (principal) square root of a complex number is nonnegative. So, we are left with the second solution set from which we get z = w₁² = 48 + 14i as the sole solution of the original equation √z − √(−z) = 6 + 8i.

Note that this is consistent with the fact that √(−z) = −i·√z = −√(−1)·√z if Im(z) > 0 since rewriting the equation as √z + i√z = 6 + 8i or √z(1 + i) = 6 + 8i or √z = (6 + 8i)/(1 + i) = 7 + i indeed gives z = (7 + i)² = 48 + 14i which has a positive imaginary part.

[FrancisZerbib]

I didn’t have time to solve this morning exercise that you posted a new one. “You are good, you. You are good!”

[dulcineafn]

Suggestion for next video: z^4+z^1/3=z^2

[RyanLewis-Johnson-wqxs]

Sqrt[z]-Sqrt[-z]=6+8i z = 48 + 14 final answer

[scottleung]

I used the second method, but forgot about the conjugate.

[DonEnsley]

problem:

√z - √-z = 6 + 8 i
√z - i √z = 6 + 8 i
√z(1 - i) = 6 + 8 i
√z(1 - i)(1+i) = (6 + 8 i)(1+i)
2√z = 6+6i+8i+8i²
= -2+14i
√z = -1 + 7 i

z = (-1 + 7 i)(-1 + 7 i)
= 1 -7i - 7i + 49 i²
= -48 - 14 i

answer
z = -48 - 14 i

[haider]

Square root of z = negative are you sure ???