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B
argw=π4
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C
|w|=√2
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D
argw=−3π4
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Solution
The correct options are A|w|=3√2 Dargw=−3π4 w=−9+3i1−2i=(−9+3i)(1+2i)(1−2i)(1+2i)=−9−18i+3i−61+4 =−15−15i5=−3−3i |w|=|−3−3i|=√(−3)2+(−3)2=√18=3√2 Now, tanα=(|−3||−3|)=1 ⇒α=π4 As w lies in third quadrant So, argw=−(π−α)=−(π−π4)=−3π4