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Tag: Solutions

i-th root of i

In my previous post we talked about i^i and we evaluated it. We found that the result is a real number, although you would expect it to be complex since i is an imaginary number. Is this the case for the i-th root of i? First of all, let’s write this as a power: Now, this is can also be written as We already know what i^i equals therefore, This means it’s simply the inverse of i^i, and so it’s a real number as well! If you use a calculator… Read more i-th root of i

i to the i-th power

Ever wondered what the imaginary unit i raised to itself is equal to? Is it going to be a complex number? Or a real one? Let’s find out! Let’s first define i: Therefore, There is an identity in maths which is in my opinion one of the most beautiful equations that exist, and that is Euler’s identity: This means At this point we have two expressions: We can use the second one to replace the -1 in the first: We no longer have the i because i times i equals… Read more i to the i-th power

Imaginary numbers | Solutions with procedure

Here are the solutions to the exercises I gave you here! Notice that cosine is an even function, therefore cos(-pi)=cos(pi). The logarithm in base 3 of 3 is 1 because 1 is the number you must raise 3 to in order to get 3. Now, why is theta raised to the log in base theta of negative i equal to negative i, i.e. the argument of the logarithm? That’s a property: Another property is the following: Hope this was useful! If you have any doubt leave a comment and I… Read more Imaginary numbers | Solutions with procedure