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If-a-b-c-are-real-numbers-and-z-is-a-complex-number-such-that-a-2-b-2-c-2-1-and-b-ic-1-a-z-then-1-iz-1-iz-equals-1-b-ic-1-ia-2-a-ib-1-c-3-1-




Question Number 20932 by Tinkutara last updated on 08/Sep/17
If a, b, c are real numbers and z is a  complex number such that, a^2  + b^2  + c^2   = 1 and b + ic = (1 + a)z, then ((1 + iz)/(1 − iz))  equals.  (1) ((b − ic)/(1 − ia))  (2) ((a + ib)/(1 + c))  (3) ((1 − c)/(a − ib))  (4) ((1 + a)/(b + ic))
$$\mathrm{If}\:{a},\:{b},\:{c}\:\mathrm{are}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{and}\:{z}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{complex}\:\mathrm{number}\:\mathrm{such}\:\mathrm{that},\:{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \\ $$$$=\:\mathrm{1}\:\mathrm{and}\:{b}\:+\:{ic}\:=\:\left(\mathrm{1}\:+\:{a}\right){z},\:\mathrm{then}\:\frac{\mathrm{1}\:+\:{iz}}{\mathrm{1}\:−\:{iz}} \\ $$$$\mathrm{equals}. \\ $$$$\left(\mathrm{1}\right)\:\frac{{b}\:−\:{ic}}{\mathrm{1}\:−\:{ia}} \\ $$$$\left(\mathrm{2}\right)\:\frac{{a}\:+\:{ib}}{\mathrm{1}\:+\:{c}} \\ $$$$\left(\mathrm{3}\right)\:\frac{\mathrm{1}\:−\:{c}}{{a}\:−\:{ib}} \\ $$$$\left(\mathrm{4}\right)\:\frac{\mathrm{1}\:+\:{a}}{{b}\:+\:{ic}} \\ $$

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