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find the modulus of the complex number 2+i/3-i
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first make the complex number with no i in the denominator
\[(2+i \div3-i) \times (3+i \div3+i)\]
ie. times by its conjugate
you get half+half(i)
then \[\sqrt{0.5+0.5^{2}}\]
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sorry thats wrong \[\sqrt{0.5^{2}+0.5^{2}}\]
and thats ur mod
(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 \[\tan \theta=1/2/1/2\] \[\tan \theta=\Pi/4\]
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