Q.If $(a+ib) = \dfrac{1+i}{1-i}$, then prove that $(a^2+b^2) = 1$.

Dividing one complex number by another nonzero complex number is done by first multiplying both the numerator and the denominator by the conjugate of the denominator, which turns the denominator into a real number (since $z\bar z=c^2+d^2$ for $z=c+id$) and lets the quotient be written back in standard $a+ib$ form. Explicitly, for $z_1=a+ib$ and $z_2=c+id\neq0$: $\dfrac{z_1}{z_2}=\dfrac{a+ib}{c+id}\times\dfrac{c-id}{c-id}=\dfrac{(ac+bd)+i(bc-ad)}{c^2+d^2}$, so the real part is $\dfrac{ac+bd}{c^2+d^2}$ and the imaginary part is $\dfrac{bc-ad}{c^2+d^2}$, both genuinely real numbers. Two special cases are used constantly: $\dfrac1i=-i$ (multiply top and bottom by $i$: $\dfrac1i=\dfrac{i}{i^2}=\dfrac{i}{-1}=-i$), and the general reciprocal $\dfrac{1}{a+ib}=\dfrac{a-ib}{a^2+b^2}$. Division also interacts predictably with the modulus and conjugate: $\left|\dfrac{z_1}{z_2 …