The argument of the complex number ( (i/2)-(2/i) ) is equal to

Question

The argument of the complex number ( (i/2)-(2/i) ) is equal to (A)  (π /4)  (B)  (3π /4)  (C)  (π /12)  (D)  (π /2)

Tardigrade Admin · Accepted Answer

Let z=(i/2)-(2/i)=(i/2)-(2i/i2) Z=(i/2)+2i=(5/2)i=0+(5/2)i arg (z)= tan -1( ( operatornameIm(z)/ operatornameRe(z)) ) = tan -1( (5/2/0) )= tan -1(∞ ) = tan -1( tan (π /2) )=(π /2)

Q. The argument of the complex number $(\frac{i}{2} - \frac{2}{i})$ is equal to

$\frac{π}{4}$

$\frac{3 π}{4}$

$\frac{π}{12}$

$\frac{π}{2}$

$\frac{3 π}{2}$

Let $z = \frac{i}{2} - \frac{2}{i} = \frac{i}{2} - \frac{2 i}{i ^{2}}$ $Z = \frac{i}{2} + 2 i = \frac{5}{2} i = 0 + \frac{5}{2} i$ $ar g (z) = tan^{- 1} (\frac{Im ( z )}{Re ( z )})$
$= tan^{- 1} (\frac{5/2}{0}) = tan^{- 1} (\infty)$
$= tan^{- 1} (tan \frac{π}{2}) = \frac{π}{2}$

Q. The argument of the complex number (2i​−i2​) is equal to

4π​

43π​

12π​

2π​

23π​

Let z=2i​−i2​=2i​−i22i​ Z=2i​+2i=25​i=0+25​i arg(z)=tan−1(Re(z)Im(z)​) =tan−1(05/2​)=tan−1(∞) =tan−1(tan2π​)=2π​