A Wonderful Math Problem. Square root i + Square root -i =?

#maths #algebra #mathtricks #complex #complexnumbers

I am Rashel 👈👉I am a Math Tutor of youtube which can be called Online Math Tutor👍What is the answer of X in this exponential Equation?
A beautiful Math Problem ❤️A Nice Algebra Problem ❤️A Exponential Equation ❤️A Beautiful Math Problem ❤️Equation Solving ❤️Rashel's Classroom ❤️
❤️Equation ❤️Math Algebra.

[Yuno_]

I just realized there are so many intelligent peoples in this comment sections👼❤

[tiemen]

Each root individually has 2 solutions, so there should be 4 solutions. If you map the solutions of root(i) and root(-i) on the complex plain (recall that by taking a root of a complex number you half the phase angle, but you can do this in either direction giving you 2 solutions). And if you then do the addition you'll easily see that all 4 solutions are in a different axis on the complex plain, and that they should be root(2), -root(2), i*root(2) and -i*root(2).

[mokouf]

I use polar form of complex number for general solution:
i = e^(2m+1/2)πi
sqrt(i) = e^(m+1/4)πi
-i = e^(2n-1/2)πi
sqrt(i) = e^(n-1/4)πi

sqrt(i) + sqrt(-i) = e^(m+1/4)πi + e^(n-1/4)πi = z
Now we have 4 cases to deal with
(m, n) is (even, even): z = e^(π/4)i + e^-(π/4)i = 1/√2 + (1/√2)i + (1/√2 - (1/√2)i) = √2
(m, n) is (even, odd): z = e^(π/4)i + e^(3π/4)i = 1/√2 + (1/√2)i + (-1/√2 + (1/√2)i) = (√2)i
(m, n) is (odd, even): z = e^(-3π/4)i + e^-(π/4)i = -1/√2 - (1/√2)i + (1/√2 - (1/√2)i) = -(√2)i
(m, n) is (odd, odd): z = e^(-3π/4)i + e^(3π/4)i = -1/√2 - (1/√2)i + (-1/√2 + (1/√2)i) = -√2

All four solutions have modulus √2 and either purely real or purely imaginary. We can easily generalize all of them.
sqrt(i) + sqrt(-i) = √2 e^(kπ/2)i where k is any integer.

[Ricardo_S]

Actually there are 4 answers, in general


The main answer is √(2)

But there is also -√(2)

and i√(2)

And also -i√(2)

±√(2) and ±i√(2)

[KarlHeinzSpock]

nice video, i learned new things about complex numbers.

[danielettedgui]

great answer. Another way is to just square the given expression and get i-i-2i^2 =2. Square root this and get +/-sqroot2!

[speakingsarcasm]

There's only one answer technically if you consider how square roots are defined in the context of complex numbers, the solution with positive real part is chosen while removing the radical... If real part is zero, the one with positive imaginary part is chosen. By chosen, I mean, it becomes the principal square roots.

In general, any given complex number can be expressed in infinitely many ways in the form r•e^(i(x+2nπ)) where n is any integer. So when we apply say square root, the number becomes r•e^(i(x/2+nπ)). We choose n such that x/2 + nπ is an acute angle, (which if not possible, our next preference is obtuse angle or the negative version of acute angle). That's about it... So there's only one square root of a complex number...

√i + √(-i)
= √(e^(iπ/2)) + √(e^(-iπ/2))
= e^(iπ/4) + e^(-iπ/4)
= 2cos(π/4)
= √2
That's the only solution if we consider only the *principal square roots* of the radicals involved.

[CipriValdezate]

A shorter way: say A=√i+√-i, then A²=(√i+√-i)²=±2, then A=±√±2, and from here the 4 solutions: √2, i√2, -√2, -i√2.

[Abitibidoug]

I found it easier to solve using polar form rather than rectangular form complex numbers.

[asc.atlantean]

√i + √-i =
[into the polar form: re^(iφ), r > 0, -π < φ ≤ π ]
= (e^(iπ/2))^(1/2) + (e^(-iπ/2))^(1/2) = e^(iπ/4) + e^(-iπ/4) =
= 2cos(π/4) = √2

no ± bs, since root functions return a single value

[KarlHeinzSpock]

i solved it "geometrically":
visualize i and minus i as arrows in the plain of complex numbers.
i has ankle 90 degrees, -i has ankle 270 degrees.
length of both arrows is 1.
interprete the operation sqrt() as division of ankle by 2.
so you get two arrows, that represent sqrt(i) and sqrt(-i).
add these two arrows as known from vector addition.
you get a triangle which leads you to the solution sqrt(2)*i,
if you apply some basic trigonometry on it.
(in triangle, use: sin(45°)=0.5sqrt(2))

[billprovince]

The sqrt bracket operator denotes the *principal" square root, hence, no need for considering the negative roots. In other words, it should have been reduced to sqrt(2)*(1 + i + 1 - i) = sqrt(2)*2 = sqrt(2). No need for +/-. If you *do* want to consider non-principal square roots, then you also need to consider the cases where you have opposite selections for the +/- on either root. Hence, sqrt(2)*((1+i)-(1-i)) = sqrt(2)*2*i = sqrt(2)*i. Similarly, another non-principal square root would be -sqrt(2)*i.

[Paul-]

As some astute commenters have pointed out, there are four solutions. The video limits solutions to the real numbers, but the problem involves complex numbers so I think they’re fair game for the solution.

You can derive SQRT(i) by setting it equal to +/- (a + bi) and doing the algebra, but that isn’t necessary. A moment’s thought should bring you to the conclusion that i multiplied by the real roots gives valid solutions because squaring i just flips the sign from +/- to -/+, and both signs are already part of the solution.

[WookieRookie]

-sqrt(2), i*sqrt(2), -I*sqrt(2) are also solutions: it depends on which branch of square root you use. Imagine solutions are when you use different branches for the root.

[blue_sand]

At 3:00 you lost the ”+/-” between (1+i) and (1-i). So, there are 2 more solutions.

[lajont]

First of all, the square root (√ ) of complex numbers are defined only by the polar form such that the argument becomes halved and the old magnitude is the square of the new magnitude. Eg √(z)=√(r)e^(it/2) given that z=re^(it). This means that the method you use might give the wrong result, but it seems like it does not.

Second, even for real numbers, √(x)>0, so even though the solution for x^2=a has two solutions, √(x) only has one solution for all x>0.

Thirdly, the function √ also only have one solution for the complex domain, where we may only choose the polar representation that gives us 0<t<2π such that z=re^(it) which then gives the uniqueness of the square root function.

Armed with this knowledge, we find that √i=e^(iπ/4)=1/√2 * (1+i) while √(-i)=e^(i3π/4)=1/√2 * (1-i). Therefore we have √i+√(-i)=1/√2 * (1+i+1-i)=2/√2=√2. This means that there is no ±, just one answer!

This relies on the fact that the square root is a function, meaning that for any input value, it has a unique output value.

[danvladoiu]

if you use z = sqrt(i) + sqrt(-i) and square z you will get the result much faster...

[ManjulaMathew-wbzn]

I find a few problems in the calculation. Sum of two unique values should result in one answer.
If your answer is back substituted
sqrt(i)+sqrt(-i)=+/-sqrt2. Squaring as is
i + 2 sort(i)sqrt(-i) - i =2
2sqrt(i)sqrt(i)sqrt(-1)=2
2i.i=2
2i^2=2
-2=2 so your answer doesn’t satisfy original equation.
isqrt2 would be the correct answer. When you back substitute LHS will simplify the same and the RHS will have an extra -i^2 term making it -2=-2.

[johnnychinstrap]

A lot of work when you can let it equal answer, A, and square both sides. ONLY TAKES 2 LINES THAT WAY. So A^2 =2 and A = +- rt2

[MikeGranby]

I just visualized the vector for i and halved the argument, and then the same for -i. The imaginary parts cancel, and the real parts add, and there’s your answer.