An imaginary number system

Do you know how the decimal system works? Well, then this should be an easy problem for you:

Why are number systems (base n n ) only defined for n N n\in \Nu ? We can easily expand this idea!

What is the decimal representation* of the following base 4 + i 2 4+i\cdot 2 number? z = 424 4 + i 2 z ={ 424 }_{ 4+i\cdot 2 } Type R e ( z ) + I m ( z ) Re\left( z \right) +Im\left( z \right) as decimal number.

*Clarification: decimal representation means, that R e ( z ) Re\left( z \right) and I m ( z ) Im\left( z \right) are decimal numbers.


The answer is 128.

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1 solution

CodeCrafter 1
Nov 10, 2019

In a number system base n n you have: x 2 x 1 x 0 = + x 2 n 2 + x 1 n 1 + x 0 n 0 \overline { \dots { x }_{ 2 }{ x }_{ 1 }{ x }_{ 0 } } =\dots +{ x }_{ 2 }\cdot { n }^{ 2 }+{ x }_{ 1 }\cdot { n }^{ 1 }+{ x }_{ 0 }\cdot { n }^{ 0 }

Therefore in base 4 + i 2 4+i⋅2 we have: z = 424 4 + i 2 = 4 ( 4 + i 2 ) 2 + 2 ( 4 + i 2 ) 1 + 4 ( 4 + i 2 ) 0 = [ 48 + i 64 ] + [ 8 + i 4 ] + [ 4 ] = 60 + i 68 \begin{aligned}z={ 424 }_{ 4+i⋅2 }&=4\cdot { \left( 4+i⋅2 \right) }^{ 2 }+2\cdot { \left( 4+i⋅2 \right) }^{ 1 }+4\cdot { \left( 4+i⋅2 \right) }^{ 0 }\\&=\left[ 48+i\cdot 64 \right] +\left[ 8+i\cdot 4 \right] +\left[ 4 \right] \\ &=60+i\cdot 68\end{aligned}

And finally: R e ( z ) + I m ( z ) = 128 \boxed { Re\left( z \right) +Im\left( z \right) =128 }

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