If (z) and (z.\overline{z} = 0) are any two complex numbers... | MATH - CONJUGATE, MODULUS AND ARGUMENT OF COMPLEX NUMBERS | SolveeIt

Question

If $z$ and $z.\overline{z} = 0$ are any two complex numbers then $z = 0$ ${Re}(z) = 0$ is equal to.

${Im}(z) = 0$

$(a + ib)(c + id)(e + if)(g + ih)$

$= A + iB,$

$(a^{2} + b^{2})(c^{2} + d^{2})(e^{2} + f^{2})(g^{2} + h^{2})$

Answer

$(a + ib)(c + id)(e + if)(g + ih)$

Explanation

Solution

$\mathbf{z =}\frac{\mathbf{1 + iy}}{\mathbf{1}\mathbf{-}\mathbf{iy}}\mathbf{=}\frac{\mathbf{1 + iy}}{\mathbf{1}\mathbf{-}\mathbf{iy}}\mathbf{\times}\frac{\mathbf{1 + iy}}{\mathbf{1 + iy}}\mathbf{=}\frac{\mathbf{(1}\mathbf{-}\mathbf{y}^{\mathbf{2}}\mathbf{) + 2iy}}{\mathbf{1 +}\mathbf{y}^{\mathbf{2}}}$

$\therefore$

$|z| = \frac{1}{1 + y^{2}}\sqrt{(1 - y^{2})^{2} + 4y^{2}} = \frac{1 + y^{2}}{1 + y^{2}} = 1$

Question: If \(z\) and \(z.\overline{z} = 0\) are any two complex numbers then \(z = 0\) \({Re}(z) = 0\) is eq...

Solution