If z -3-2- i -2- Then - z will beA- j -6-C- -5 -6

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 =√3/2+ i...

Question

If $z = \frac{\sqrt{3}}{2} + \frac{i}{2},$ Then $arg (- z)$ will be

- \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 $- \frac{5 π}{6}$
Now $- z = - \frac{\sqrt{3}}{2} - \frac{i}{2}$

We have $tan α = \frac{1}{\sqrt{3}}$
$\Rightarrow α = \frac{π}{6}$
Since the complex number $- z$ lies in the third quadrant, so
$arg (- z) = θ = - (π - α)$
$= - (π - \frac{π}{6}) = - \frac{5 π}{6}$

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Geometrical Representation of a Complex Number

Standard XII Mathematics

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If z=√32+i2, Then arg(−z) will be

The correct option is B −5π6Now −z=−√32−i2 We have tanα=1√3 ⇒α=π6 Since the complex number −z lies in the third quadrant, so arg(−z)=θ=−(π−α) =−(π−π6)=−5π6

If $z = \frac{\sqrt{3}}{2} + \frac{i}{2},$ Then $arg (- z)$ will be

The correct option is B $- \frac{5 π}{6}$
Now $- z = - \frac{\sqrt{3}}{2} - \frac{i}{2}$

We have $tan α = \frac{1}{\sqrt{3}}$
$\Rightarrow α = \frac{π}{6}$
Since the complex number $- z$ lies in the third quadrant, so
$arg (- z) = θ = - (π - α)$
$= - (π - \frac{π}{6}) = - \frac{5 π}{6}$