নতি (Argument)
1−(11−11+i) 1 - \left ( \frac{1}{1 - \frac{1}{1 + i}} \right ) 1−(1−1+i11) এর মডুলাস ও আর্গুমেন্ট-
1,0
1,π
1−11−11+i=1−11+i−11+i=1−1+ii=i−1−ii=i2i=i∴ মডুলाস =1 এবং আর্গুমেন্ট, θ=tan−1(10)=tan−1tan(π2)=π2 \begin{array}{l}\text { } 1-\frac{1}{1-\frac{1}{1+i}}=1-\frac{1}{\frac{1+i-1}{1+i}} \\ =1-\frac{1+i}{i}=\frac{i-1-i}{i}=\frac{i^{2}}{i}=i \\ \therefore \text { মডুলाস }=1 \text { এবং আর্গুমেন্ট, } \theta=\tan ^{-1}\left(\frac{1}{0}\right) \\ =\tan ^{-1} \tan \left(\frac{\pi}{2}\right)=\frac{\pi}{2}\end{array} 1−1−1+i11=1−1+i1+i−11=1−i1+i=ii−1−i=ii2=i∴ মডুলाস =1 এবং আর্গুমেন্ট, θ=tan−1(01)=tan−1tan(2π)=2π
Find the value of θ\thetaθ if (3+2isinθ)(1−2isinθ )\frac{\left(3+2i\sin\theta\right)}{\left(1-2i\sin\theta\ \right)}(1−2isinθ )(3+2isinθ) Is purely imaginary.
The modulus of the complex number zzz such that ∣z+3−i∣=1\left| z + 3 - i\right | = 1∣z+3−i∣=1 and argz=π\arg{z} = \piargz=π is equal to
Argument of the complex number (−1−3i2+i).\left( \dfrac { - 1 - 3 i } { 2 + i } \right).(2+i−1−3i).
z1=−1−i3,z2=3−iz_{1}=-1-i \sqrt{3}, z_{2}=\sqrt{3}-iz1=−1−i3,z2=3−i.
