\( i ^ { 105 } : ( i ^ { 10 } , i ^ { 29 } ) = \)

To solve \( i^{105} : (i^{10}, i^{29}) \), we first find \( i^{105} \), \( i^{10} \), and \( i^{29} \). 1. The powers of \( i \): - \( i^1 = i \) - \( i^2 = -1 \) - \( i^3 = -i \) - \( i^4 = 1 \) - \( i^5 = i \) (and it repeats every 4) Now, we calculate \( i^{105} \): - \( 105 \mod 4 = 1 \) (since 105 divided by 4 gives a remainder of 1) - So, \( i^{105} = i^1 = i \). Next, we calculate \( i^{10} \) and \( i^{29} \): - \( 10 \mod 4 = 2 \), therefore \( i^{10} = i^2 = -1 \). - \( 29 \mod 4 = 1 \), thus \( i^{29} = i^1 = i \). Now we substitute these values into the expression: - The expression \( i^{105} : (i^{10}, i^{29}) \) becomes \( i : (-1, i) \). To find the ratio \( i : (-1, i) \): - This can be understood as how many times the first term (\( i \)) goes into each of the terms in the list: - \( i : -1 \) results in \( -i \) (or \( (-1, 1) \) if you write both in complex form). - \( i : i \) is simply 1. Therefore, the final results are: - For \( i : -1 \) you can think of it as \( -i \) which in polar form is \( 1 \) at \( 270^\circ \) or \( \frac{3\pi}{2} \). - For \( i : i \) is literally \( 1 \). This means the complete answer remains in a descriptive format related to \( i \) and includes: - In complex division, each outcome echoes the relationship among the complex numbers.