" If "a+ib=(c+i)/(c-i)," where "c" is real,then "(b)/(a)=

If a+ib=(c+i)/(c-i), where c is real,prove that: a^(2)+b^(2)=1 and (b)/(a)=(2c)/(c^(2)-1)

If a+ib= (c+i)/(c-i) , where c is real, prove that a^2+b^2=1 and b/a= (2c)/(c^2-1) .

If a+i b=(c+i)/(c-i) , where c is real, prove that: a^2+b^2=1 and b/a=(2c)/(c^2-1)dot

If the modulus of the complex number a+ib (b neo) is 1, showthat the complex number can be represented as follows : a+ib=(c+i)/(c-i) , where c is a real quantity.

If (c+i)/(c-i)= a+ ib , where a, b, c are real, then a^2+ b^2 equals :

If z=a+ib, |z|=1 and b ne 0, show that z can be represented as z=(c+i)/(c-i) where c is a real number.

If one root of z^2 + (a + i)z+ b +ic =0 is real, where a, b, c in R , then c^2 + b-ac=

Evaluate |(a+ib,c+id),(-c+id,a-ib)| where i=sqrt(-1)

The value of Arg [i ln ((a-ib)/(a+ib))] , where a and b are real numbers, is

If (1 + 2i) is a root of the equation x^(2) + bx + c = 0 . Where b and c are real, then (b,c) is given by :