Question

Find the modulus, argument and the principal argument of the complex numbers.

$(tan1-i{)}^2$

• A. $Modulus={\sec }^{2}1,Arg\left(z\right)=2n\pi +(2-\pi ),PrincipalArg\left(z\right)=(2-\pi )$
• B. $Modulus={\text{cosec}}^{2}1,Arg\left(z\right)=2n\pi -(2-\pi ),PrincipalArg\left(z\right)=(-2-\pi )$
• C. $Modulus={\sec }^{2}1,Arg\left(z\right)=2n\pi -(2-\pi ),PrincipalArg\left(z\right)=(-2-\pi )$
• D. $Modulus={\text{cosec}}^{2}1,Arg\left(z\right)=2n\pi +(2-\pi ),PrincipalArg\left(z\right)=(2-\pi )$

Answer 1

The correct option is A $Modulus={\sec }^{2}1,Arg\left(z\right)=2n\pi +(2-\pi ),PrincipalArg\left(z\right)=(2-\pi )$

Let
$z=\tan 1-i$
Then
$|z|=\sqrt{{\tan }^{2}1+1}=\sec 1$
$Arg\left(z\right)={\tan }^{-1}\left(\frac{-1}{\tan 1}\right)$
$=-{\tan }^{-1}\left(cot\right(1\left)\right)$
$=-(\frac{\pi }{2}-{\cot }^{-1}(cot\left(1\right)\left)\right)$
$=1-\frac{\pi }{2}$
Hence
$Z=(\tan 1-i{)}^2=z^2=|z{|}^2.e^{2i.(1-\frac{\pi }{2})}$
$={\sec }^2\left(1\right)e^{i.(2-\pi )}$
Hence
$|Z|={\sec }^2\left(1\right)$ and principal Arg(z)=$(2-\pi )$.