Fun with j: who can solve these exactly WITHOUT a calculator?

Question

Let j = sqrt(-1) (use "i" if you prefer math-guys)

Put these in rectangular form:

1) j^j

2) sqrt(j)

3) cuberoot(j) e.g. j^(1/3)

4) cos(x) = 2 (solve for x)

Don't use a computer or either - I mean, solve "by hand."

I will post my answers in an hour.

OK -

j = sqrt(-1) = exp(j*pi/2) (Euler)

Then

1) j^j = exp(-pi/2)

2) sqrt(j) = exp(j*pi/2)^1/2 = exp(j*pi/4) = j*sin(pi/4)

3) same, except j*sin(pi/6)

4) complicated -

but cos(x) = (exp(jx) + exp(-jx))/2 = 2

let a = exp(jx), then solve for a.

a = (4 (+ or -) sqrt(16 - 4*j))/2 = exp(jx)

solve that for x, you got it.

cos(x) varies 1 to -1 only for x, real.

ooops - one 2) and 3) - I mean

exp(j*pi/4) = cos(pi/4) + j*sin(pi/4)

exp(j*pi/6) = cos(pi/6) + j*sin(pi/6)

Answer 1

Wow... no one else got it? Is that really the best Y!A has to offer? Here's the answer and solution method:

#1: i^i = (e^(ln i))^i = (e^(i(π/2±2πk)))^i = e^(i*i(π/2±2πk)) = e^(-π/2±2πk) = 1/√(e^π) * e^(2πk), where k is any integer.

The only step there that requires explanation is ln i = iπ/2, which comes from euler's formula: e^(ix) = cos x + i sin x. Obviously, this means that cos x must be zero and sin x must be 1, which only occurs at π/2±2πk (the 2πk reflects the periodic nature of the trigonometric functions). Everything else is just simple algebra.

#2: ±(√2 + i √2)/2 - this one is fairly obvious, if you know the geometric interpretation of complex exponentiation and the sin of π/4, and of course verifying the answer doesn't even require you to know that.

#3: (√3 + i)/2, (-√3 + i)/2, -i -- knowing that the cube roots of a number are spaced around the unit circle at angles of 2π/3 is very helpful here.

#4: Okay, admittedly I had to look this one up - I don't usually work with trig functions in complex numbers. Anyway, the key here is that cos (ix) = cosh (x), and arccosh (x) = ln (x ± √(x²-1)).
Thus cos x = 2 → cosh (-ix) = 2 → -ix = ln (2 ± √3) → x = i ln (2 ± √3).

Answer 2

You can't square root a negative number, as a minus times a minus equals a plus.

Instead, I will make -j = sqrt(1)

1. -1^-1 = 1

2. You won't be able to do this one as, in order to make j equal something I had to make -j a plus, this makes j a minus and you can't square a minus.

3. Again, won't be able to do this because of the above reason.

4. Cos(x) ranges from -1 to 1.

Answer 3

1) =-1 -1^(2/2)
2) ? -1^(1/4)
3) -1^(1/6)
4) no solution as cos x ranges from -1 to +1

Answer 4

1) -1

2) sqrt(2)/2 + j*sqrt(2)/2

3) sqrt(3)/2 + j*1/2

4) x = -i ln (2-sqrt(3))

Answer 5

there is no way I can solve that by hand or computer. This is totally above my math skills.

Answer 6

I'm guessing 0.

Answer 7

-1
cant sq root a neg
-.7 or something
1

Answer 8

I can.

TFTP