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



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priyab2
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first make the complex number with no i in the denominator

priyab2
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\[(2+i \div3i) \times (3+i \div3+i)\]

priyab2
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ie. times by its conjugate

priyab2
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you get half+half(i)

priyab2
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then \[\sqrt{0.5+0.5^{2}}\]

priyab2
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sorry thats wrong \[\sqrt{0.5^{2}+0.5^{2}}\]

priyab2
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and thats ur mod

adhirajmajumder
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(2+i)/(3i)
=(2+i)(3+i)/(3+i)(3i)
=(6+5i1)/9+1
=(5+5i)/10
=(1+i)/2
IZI=\[\sqrt{(1/2)^2+(1/2)^2}\]
=\[\sqrt{.25+.25}\]
=\[\sqrt{.50}\]
=0.7071
\[\tan \theta=1/2/1/2\]
\[\tan \theta=\Pi/4\]