If z - -32 - i2- then -z is

Given: z = √(3)2 + i2 Now, z = √(3)2 i2 tanα nbsp; = 12 √(3)2= 1√(3) ⇒α= π6 Since the complex number z lies in the third quadrant, so ( z) = (π α) ...

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Standard XII

Mathematics

Geometrical Representation of a Complex Number

If z = √32 + ...

Question

If $z = \frac{\sqrt{3}}{2} + \frac{i}{2},$ then $arg (- z)$ is

- \frac{π}{6}

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- \frac{5 π}{6}

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\frac{5 π}{6}

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\frac{π}{6}

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Solution

The correct option is B $- \frac{5 π}{6}$
Given $z = \frac{\sqrt{3}}{2} + \frac{i}{2}$
Now,
$- z = - \frac{\sqrt{3}}{2} - \frac{i}{2} tan α = \frac{1}{\sqrt{3}} \Rightarrow α = \frac{π}{6}$

Since the complex number $- z$ lies in the third quadrant, so
$arg (- z) = - (π - α) = - (π - \frac{π}{6}) = - \frac{5 π}{6}$

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Geometrical Representation of a Complex Number

Standard XII Mathematics

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If z=√32+i2, then arg(−z) is

The correct option is B −5π6Given z=√32+i2 Now, −z=−√32−i2tanα=1√3⇒α=π6 Since the complex number −z lies in the third quadrant, so arg(−z)=−(π−α) =−(π−π6)=−5π6

If $z = \frac{\sqrt{3}}{2} + \frac{i}{2},$ then $arg (- z)$ is