Question

Find the modulus and the argument of the complex number

z = - 1 - i \sqrt{3}

.

Accepted Answer

Given: $z = - 1 - i \sqrt{3}$

Let $z = r (\cos θ + i \sin θ)$

$\Rightarrow - 1 - i \sqrt{3} = r \cos θ + i r \sin θ$

Here, $r$ is modulus of a complex number and $θ$ is argument of complex number.

On comparing real and imaginary parts we have,

Let $r \cos θ = - 1$ and $r \sin θ = - \sqrt{3}$ .................. $(i)$

On squaring and adding, we obtain

$(r \cos θ)^{2} + (r \sin θ)^{2} = (- 1)^{2} + (- \sqrt{3})^{2}$

$\Rightarrow r^{2} (\cos^{2} θ + \sin^{2} θ) = 1 + 3$

$\Rightarrow r^{2} = 4 [\cos^{2} θ + \sin^{2} θ = 1]$

$\Rightarrow r = \sqrt{4} = 2$ (Since, $r$ is always positive)

$∴$ Modulus $= 2$

For argument of the complex number:

$2 \cos θ = - 1$ and $2 \sin θ = - \sqrt{3}$ ( from (i))

$\Rightarrow \cos θ = \frac{- 1}{2}$ and $\sin θ = \frac{- \sqrt{3}}{2}$

Since both the values of $\sin θ$ and $\cos θ$ are negative and $\sin θ$ and $\cos θ$ are negative in III quadrant,

$∴$ Argument $= - (π - \frac{π}{3}) = \frac{- 2 π}{3}$

Thus, the modulus and argument of the complex number $- 1 - \sqrt{3} i$ are $2$ and $\frac{- 2 π}{3}$ respectively.

EXERCISE 5.2

Jharkhand Board Mathematics Solutions for EXERCISE 5.2

Questions from EXERCISE 5.2 with Hints & Solutions

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