Complex Number Solution Question (1 Viewer)

pikachu975 · Dec 7, 2016

http://prntscr.com/dg0pux

z1 = root(3) + i
z2 = -root(2) + root(2)i

Part i was to express them both in mod arg form
z1 = 2cispi/6
z2 = 2cis3pi/4

How did they get from -arg(z1+z2) to -(argz1 + argz2) / 2?

Thanks!

InteGrand · Dec 7, 2016

pikachu975 said:
http://prntscr.com/dg0pux

z1 = root(3) + i
z2 = -root(2) + root(2)i

Part i was to express them both in mod arg form
z1 = 2cispi/6
z2 = 2cis3pi/4

How did they get from -arg(z1+z2) to -(argz1 + argz2) / 2?

Thanks!

$\noindent It's because $z_{1}$ and $z_{2}$ have the same length (modulus), so $z_{1} + z_{2}$ forms the diagonal of a \emph{rhombus} (if they weren't the same length, it wouldn't be a rhombus, just a regular parallelogram. But since they're the same length, it's a special parallelogram, namely a rhombus.).$

$\noindent Since $z_{1} + z_{2}$ corresponds to a diagonal of this rhombus and we know that \emph{diagonals of a rhombus bisect its angles}, the vector for $z_{1} + z_{2}$ must lie halfway between $z_{1}$ and $z_{2}$ (in terms of angle). In other words, the argument of $z_{1}+z_{2}$ is halfway between those of $z_{1}$ and $z_{2}$, i.e. $\arg \left(z_{1} + z_{2}\right) = \frac{1}{2}\left(\arg z_{1} + \arg z_{2}\right)$.$

Complex Number Solution Question (1 Viewer)

pikachu975

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InteGrand

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