Q1: What is the complex conjugate of $z$ in Cartesian form?
A1: $z^* = a - jb$
Q2: What is the complex conjugate of $z$ in polar form?
A2: Note that $z^*$ is simply the reflection of $z$ over the real axis of the complex plane. It then follows that:
1.The magnitude of $z^*$ is the same as the magnitude of $z$
2. The angle that $z^*$ makes with respect to the positive real axis is the negative value of the angle that $z$ makes with the positive real axis.
Hence, $z^*= |z|e^{-j\theta}$
Q3: How can you simplify the expression $zz^*$?
A3: Although the multiplication suggests that you use polar form to simplify the product, Cartesian form yields an interesting result.
$zz^*=(a+jb)(a-jb)$
Notice that the cross-terms (i.e. the O and I products of FOIL) negate each other.
$zz^*=a^2+b^2$
$zz^*={|z|}^2$
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