Question

find the modulus of the complex number 2+i/3-i

Answer 1

first make the complex number with no i in the denominator

Answer 2

\[(2+i \div3-i) \times (3+i \div3+i)\]

Answer 3

ie. times by its conjugate

Answer 4

you get half+half(i)

Answer 5

then \[\sqrt{0.5+0.5^{2}}\]

Answer 6

sorry thats wrong \[\sqrt{0.5^{2}+0.5^{2}}\]

Answer 7

and thats ur mod

Answer 8

(2+i)/(3-i)
=(2+i)(3+i)/(3+i)(3-i)
=(6+5i-1)/9+1
=(5+5i)/10
=(1+i)/2

IZI=\[\sqrt{(1/2)^2+(1/2)^2}\]
=\[\sqrt{.25+.25}\]
=\[\sqrt{.50}\]
=0.7071
\[\tan \theta=1/2/1/2\]
\[\tan \theta=\Pi/4\]