B bluemath New member Joined Sep 20, 2015 Messages 43 Nov 8, 2015 #1 Hello, Just for correction please : We have z = -i(eix) x is the argument, but what's about the module of z ? The module is equal to 1 (because we argue in term of distance with -i) Right ? Thanks
Hello, Just for correction please : We have z = -i(eix) x is the argument, but what's about the module of z ? The module is equal to 1 (because we argue in term of distance with -i) Right ? Thanks
pka Elite Member Joined Jan 29, 2005 Messages 11,978 Nov 8, 2015 #2 bluemath said: Hello, Just for correction please : We have z = -i(eix) x is the argument, but what's about the module of z ? The module is equal to 1 (because we argue in term of distance with -i) Right ? Click to expand... YES! Here are facts to remember. \(\displaystyle |z\cdot w|=|z|\cdot|w|~\&~(\forall\theta )\left(|e^{i\theta}\right)|=1\).
bluemath said: Hello, Just for correction please : We have z = -i(eix) x is the argument, but what's about the module of z ? The module is equal to 1 (because we argue in term of distance with -i) Right ? Click to expand... YES! Here are facts to remember. \(\displaystyle |z\cdot w|=|z|\cdot|w|~\&~(\forall\theta )\left(|e^{i\theta}\right)|=1\).
