`a`, `b`,`c` are three complex numbers on the unit circle `|z|=1`, such that `abc=a+b+c`. Then `|ab+bc+ca|` is equal to
`a`, `b`,`c` are three complex numbers on the unit circle `|z|=1`, such that `abc=a+b+c`. Then `|ab+bc+ca|` is equal to
A. `3`
B. `6`
C. `1`
D. `2`
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Correct Answer - C
`(c )` `abara=bbarb=cbarc=1`
`:.bara=(1)/(a)` etc.
Now, `|abc|=|a+b+c|`
`=|bara+barb+barc|`
`=|(1)/(a)+(1)/(b)+(1)/(c )|`
`=|(sumab)/(abc)|`
`:.|sumab|=|abc||abc|=(|a||b||c|)^(2)=1`
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