(3i/√2-i)^2的虚部是多少?是(根号2)-i

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(3i/√2-i)^2的虚部是多少?是(根号2)-i
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(3i/√2-i)^2的虚部是多少?是(根号2)-i
(3i/√2-i)^2的虚部是多少?
是(根号2)-i

(3i/√2-i)^2的虚部是多少?是(根号2)-i
原式=(3i)²/(√2-i)²
=-9/(1-2√2i)
=-9(1+2√2i)/(1-2√2i)(1+2√2i)
=-9(1+2√2i)/(1+8)
=-1-2√2i
所以虚部是-2√2

(3i/√2-i)^2
=(-9)/(2-1-2√2i)
=-9/(1-2√2i)
=-9(1+2√2i)/[(1-2√2i)(1+2√2i)]
=-9(1+2√2i)/(1+8)
=-1-2√2i
虚部是-2√2