If (0 + i0)then (1 + 0i) is equal to. | MATH - INTEGRAL POWER OF IOTA, ALGEBRAIC OPERATIONS AND EQUALITY OF COMPLEX NUMBERS | SolveeIt

If $0 + i0$ then $1 + 0i$ is equal to.

$0 + i$

$1 + i$

$a^{2} + b^{2} = 1,$

$\frac{1 + b + ia}{1 + b - ia} =$

Answer

$\frac{1 + b + ia}{1 + b - ia} =$

Explanation

$x = 0$

$24x^{2} + 9ixy - 6y^{2} + 16ixy = 75i - 30$

∴ ${Re}(z) = 0 \Rightarrow {Im}(z^{2}) = 0$

$\frac{5( - 8 + 6i)}{(1 + i)^{2}} = a + ib$

$\frac{- 40 + 30\ i}{2i} = 15 + 20\ i = a + ib$

$a = 15$ .

Question: If \(0 + i0\)then \(1 + 0i\) is equal to....