If Z_1 = 4i^{40} - 5i^{35} + 6i^{17} + 2 and | vedclass

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(B) Given $Z_1 = 4i^{40} - 5i^{35} + 6i^{17} + 2$.Since $i^4 = 1$,$i^{40} = (i^4)^{10} = 1$.$i^{35} = i^{32} 	imes i^3 = 1 	imes (-i) = -i$.$i^{17}

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(B) Given $Z_1 = 4i^{40} - 5i^{35} + 6i^{17} + 2$.Since $i^4 = 1$,$i^{40} = (i^4)^{10} = 1$.$i^{35} = i^{32} 	imes i^3 = 1 	imes (-i) = -i$.$i^{17} = i^{16} 	imes i = 1 	imes i = i$.Substituting these values: $Z_1 = 4(1) - 5(-i) + 6(i) + 2 = 4 + 5i + 6i + 2 = 6 + 11i$.Given $Z_2 = -1 + i$.$Z_1 + Z_2 = (6 + 11i) + (-1 + i) = (6 - 1) + (11i + i) = 5 + 12i$.The modulus $|Z_1 + Z_2| = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$.

If $Z_1 = 4i^{40} - 5i^{35} + 6i^{17} + 2$ and $Z_2 = -1 + i$,where $i = \sqrt{-1}$,then $|Z_1 + Z_2| = $

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