- #1

I am starting imaginary numbers in school and I wondered, what is x=i

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- Thread starter tomas
- Start date

- #1

I am starting imaginary numbers in school and I wondered, what is x=i^{i}

- #2

e^{-.5[pi]}

Do you know taylor series? If not, learn them, you won't understand the answer until you do.

-quick edit

Do you know taylor series? If not, learn them, you won't understand the answer until you do.

-quick edit

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- #3

Lonewolf

- 337

- 1

x=i

Taking logs to base e

ln(x) = i*ln(i)

Since ln(i)= i*Pi/2, ln(x)=-Pi/2, so raising both sides to the power of e we obtain.

x = exp(-Pi/2)

Explaination of ln(i) = -i*Pi/2

if w is a complex number such that exp(w) = z, then w = ln|z|+i*arg(z). Since |i|=1, and arg(i)=Pi/2, ln(i) turns out to be -Pi/2.

- #4

Antiproton

- 39

- 0

x=i^i => x = exp(i*log(i))

This is a definition from Complex Analysis

log(i) = Log|i| + i*arg(i)

This is the definition of log(z) where z is complex. Note it is not quite the same as ln(x).

Log(x) is a real function and works exactly the same as ln(x), its just a different terminology used in math.

arg(i) is definited as the angle the 'imaginary vector' makes with the positive real axis.

|i| is the magnitude of i (absolute value), which is 1.

So, log(i) = Log(1) + i*arg(i) = 0 + i*Pi/2 + i*2*Pi*k where k is an integer (by definition of arg(z))

Thus,x = i^i = exp(i*i(Pi/2 + 2*Pi*k) = exp(-Pi/2 - 2*Pi*k) k element of

Which provides for an infinite number of solutions.

Hope that helps.

- #5

Lonewolf

- 337

- 1

- #6

Antiproton

- 39

- 0

- #7

Lonewolf

- 337

- 1

Fair enough. That makes sense.

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