(C) Given expression: $i^{18}-3i^7+i^2(1+i^4)(i)^{22}$ We know that $i^2 = -1$,$i^3 = -i$,$i^4 = 1$. $i^{18} = (i^4)^4 \times i^2 = (1)^4 \times (-1)

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(C) Given expression: $i^{18}-3i^7+i^2(1+i^4)(i)^{22}$ We know that $i^2 = -1$,$i^3 = -i$,$i^4 = 1$. $i^{18} = (i^4)^4 \times i^2 = (1)^4 \times (-1) = -1$ $i^7 = (i^4) \times i^3 = 1 \times (-i) = -i$ $i^{22} = (i^4)^5 \times i^2 = (1)^5 \times (-1) = -1$ Substituting these values: $-1 - 3(-i) + (-1)(1+1)(-1)$ $= -1 + 3i + (-1)(2)(-1)$ $= -1 + 3i + 2$ $= 1 + 3i$

Class 11 Mathematics 4-1.Complex numbers Integral power of iota, Algebraic operations and Equality of complex numbers

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By simplifying $i^{18}-3i^7+i^2(1+i^4)(i)^{22}$,we get

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$-1+3i$
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$1-3i$
C
$1+3i$
D
$-1-3i$

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