নতি (Argument)
যদি z1=1−i,z2=3+i z_{1} = 1 - i , z_{2} = \sqrt{3} + i z1=1−i,z2=3+i হয় তবে z2z1 \frac{z_{2}}{z_{1}} z1z2 এর নতি --
θ1=tan−1−11=tan−1{tan(−π4)}=−π4 \begin{array}{l} \theta_{1}=\tan ^{-1} \frac{-1}{1} \\ =\tan ^{-1}\left\{\tan \left(-\frac{\pi}{4}\right)\right\}=-\frac{\pi}{4} \end{array} θ1=tan−11−1=tan−1{tan(−4π)}=−4π
θ2=tan−113=tan−1(tanπ6)=π6∴z2z1 এর নতি =θ2−θ1=π6−(−π4)=2π+3π12=5π12 \begin{array}{l}\theta_{2}=\tan ^{-1} \frac{1}{\sqrt{3}}=\tan ^{-1}\left(\tan \frac{\pi}{6}\right)=\frac{\pi}{6} \\ \therefore \frac{z_{2}}{z_{1}} \text { এর নতি } \\ =\theta_{2}-\theta_{1}=\frac{\pi}{6}-\left(\frac{-\pi}{4}\right)=\frac{2 \pi+3 \pi}{12}=\frac{5 \pi}{12}\end{array} θ2=tan−131=tan−1(tan6π)=6π∴z1z2 এর নতি =θ2−θ1=6π−(4−π)=122π+3π=125π
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).
If z=20i−21+21+20iz=\sqrt{20i-21}+\sqrt{21+20i}z=20i−21+21+20i, then the principal value of arg 'z' can be
