If $2 i^{2} + 6 i^{3} + 3 i^{16} - 6 i^{19} + 4 i^{25} = x + i y$ , then $(x, y) =$

(1, 4)

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(4, 1)

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(- 1, 4)

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(- 1, - 4)

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Solution

The correct option is A $(1, 4)$
$i^{4 n} = 1; i^{4 n + 1} = + i; i^{4 n + 2} = - 1; i^{4 n + 3} = - i$
$∴ 2 i^{2} + 6 i^{3} + 3 i^{16} - 6 i^{19} + 4 i^{25}$
$= 2 (- 1) + 6 (- i) + 3 (1) - 6 (- i) + 4 (i)$
$= - 2 + 3 - 6 i + 6 i + 4 i$
$= 1 + 4 i$
$(x, y) = (1, 4)$

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If 2i2+6i3+3i16−6i19+4i25=x+iy, then (x,y)=

The correct option is A (1,4)i4n=1;i4n+1=+i;i4n+2=−1;i4n+3=−i∴2i2+6i3+3i16−6i19+4i25=2(−1)+6(−i)+3(1)−6(−i)+4(i)=−2+3−6i+6i+4i=1+4i(x,y)=(1,4)

If $2 i^{2} + 6 i^{3} + 3 i^{16} - 6 i^{19} + 4 i^{25} = x + i y$ , then $(x, y) =$

The correct option is A $(1, 4)$
$i^{4 n} = 1; i^{4 n + 1} = + i; i^{4 n + 2} = - 1; i^{4 n + 3} = - i$
$∴ 2 i^{2} + 6 i^{3} + 3 i^{16} - 6 i^{19} + 4 i^{25}$
$= 2 (- 1) + 6 (- i) + 3 (1) - 6 (- i) + 4 (i)$
$= - 2 + 3 - 6 i + 6 i + 4 i$
$= 1 + 4 i$
$(x, y) = (1, 4)$