The argument of 1- i -1- i isA- -2B- 3 -2C- -2D- 5 -2

The correct option is C π/2 Let z = 1 i/1+i ⇒ z=1 i/1+i×1 i/1 i ⇒ z=1+i^2 2i/1 i^2 ⇒ z=1 1 2i/1+1 ⇒ z= 2i/2 ⇒ z= i Since, z lies on negative direction of ...

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Standard XII

Mathematics

Properties of Argument

The argument ...

Question

The argument of $\frac{1 - i}{1 + i}$ is

$\frac{- π}{2}$

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$\frac{π}{2}$

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$\frac{3 π}{2}$

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$\frac{5 π}{2}$

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Solution

The correct option is A
$\frac{- π}{2}$

$Let z = \frac{1 - i}{1 + i} \Rightarrow z = \frac{1 - i}{1 + i} \times \frac{1 - i}{1 - i} \Rightarrow z = \frac{1 + i^{2} - 2 i}{1 - i^{2}} \Rightarrow z = \frac{1 - 1 - 2 i}{1 + 1} \Rightarrow z = \frac{- 2 i}{2} \Rightarrow z = - i$

Since, z lies on negative direction of imaginary axis.

Therefore, arg (z) = $\frac{- π}{2}$

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The argument of 1−i1+i is

The correct option is A −π2 Let z = 1−i1+i⇒ z=1−i1+i×1−i1−i⇒ z=1+i2−2i1−i2⇒ z=1−1−2i1+1⇒ z=−2i2 ⇒ z=−i Since, z lies on negative direction of imaginary axis. Therefore, arg (z) = −π2

The argument of $\frac{1 - i}{1 + i}$ is