Too complex to solve!

Algebra Level 4

Let $z = i^{\lceil i^{i} \rceil}$ , where $i = \sqrt {-1}$ .

If $k = Re(z)+Im(z)$ , find $\lfloor 100k \rfloor$ .

The answer is 100.000.

1 solution

Jake Lai
Feb 28, 2015

It is known that

$i^{i} = (e^{i\pi/2})^{i} = \frac{1}{\sqrt{e^{\pi}}} \in (0,1)$

Thus, $z = i^{\lceil i^{i} \rceil} = i^{1} = i$ . Since $k = Re(z)+Im(z) = 1$ , $\lfloor 100k \rfloor = \boxed{100}$ .

For some reason, I can't tag Utkarsh Bansal.

In future problems, please write your plaintext in, well, plaintext, rather than wrapping everything in $\setminus ( \ \cdots \ \setminus )$ . For one, it looks way better; you don't have to put \quad between every word either.

Also, it isn't necessary to type { a }^{ b } when trying to get $a^{b}$ . Just a^{b} will do. Same goes for subscript.

Thanks for posting your problems here. I hope you'll continue to enjoy Brilliant and our community!

Jake Lai - 6 years, 3 months ago

When you reach the extent of laziness that I have, you'll stop using braces for single digit exponents/subscript values.

Example: a^b also gives $a^b$ . You can check by hovering over the $\LaTeX$ code with your mouse pointer.

Prasun Biswas - 6 years, 3 months ago

Thank you for your suggestion

Utkarsh Bansal - 6 years, 3 months ago

Too complex to solve!

The answer is 100.000.

1 solution

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