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Question

If (1+i)(2+3i)(34i)(23i)(1i)(3+4i)=a+ib, then a2+b2=

A
132
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B
25
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C
144
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D
128
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E
1
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Solution

The correct option is E 1
Given, (1+i)(2+3i)(34i)(23i)(1i)(3+4i)=a+ib, then a2+b2
(1+i)(2+3i)(34i)(23i)(1i)(3+4i)=|a+ib|
1+14+99+164+91+19+16=a2+b2
Squaring both sides we get
2×13×2513×2×25=a2+b2
a2+b2=1

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