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Q.

If Z=1                     12i         3+5i1+2i         5                 10i35i        10i               11 then 

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a

Z is purely real

b

arg Z=π4

c

Z is purely imaginary

d

Z = 0

answer is B.

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Detailed Solution

1            12i           3+5i1+2i    5                10i35i  10i              11z=155+100i212i11+22i30i+50i2         +3+5i10i20i2+1525iz=5510012i398i+3+5i3535i z=155+39+8i78i+16+105105i+175i+175 z=180+oi

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