Question

convert -3 into its polar form.

Answer should be:
◍Written step by step
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Answer 1

Complex Numbers

Complex number is a number that can be expressed in the form of $a + ib$, where $a$ and $b$ are real numbers and $i$ is an imaginary number called iota.

The length of the line segment is called modulus. The modulus of the complex number $z = a + ib$ is $|z| = r = \sqrt{a^2 + b^2}$.

Where $a = r \cos(\theta)$ and $b= r \sin(\theta)$

The polar form of the complex number $z = a + ib$ is $z = r \cos(\theta) + ir \sin(\theta)$.

Coming to the question now, We are asked to convert $-3$ into polar form.

Let $r \cos(\theta) + ir \sin(\theta) = -3 + 0i$. Therefore $r \cos(\theta) = -3$ and $r \sin(\theta) = 0$, where $r > 0$ and $-\pi < \theta \leqslant \pi$.

Now on squaring and adding, we get:

$\implies r^2 \cos^2(\theta) + r^2 \sin^2(\theta) = (-3)^2 + (0)^2 \\ \\ \implies r^2 \big(\cos^2(\theta) + \sin^2(\theta) \big) = (-3)^2 + (0)^2 \\ \\ \implies r^2 \big(\cos^2(\theta) + \sin^2(\theta) \big) = 9 + 0 \\ \\ \implies r^2 \big(\cos^2(\theta) + \sin^2(\theta) \big) = 9 \\ \\ \implies {r}^{2} = 9 \\ \\ \implies r = \sqrt{9} \\ \\ \implies r = 3 \qquad [r > 0]$

Therefore,

$\implies r \big(\cos(\theta)\big) = -3 \\ \\ \implies 3\big(\cos(\theta)\big) = -3 \\ \\ \implies \cos(\theta) = -1$

And,

$\implies r \big(\sin(\theta)\big) = 0 \\ \\ \implies 3\big(\sin(\theta)\big) = -3 \\ \\ \implies \sin(\theta) = 0$

Since $\cos(\theta)$ is -ve and $\sin(\theta)$ is +ve, it means that $\theta$ lies in second quadrant.

Therefore, $\cos(\theta) = \cos \pi$ and $\sin(\theta) = \sin \pi$. So, $\theta = \pi$ and $r = 3$.

Hence, the required polar form of -3 is r(cosθ + i sinθ) i.e., 3(cosθ + i sinθ).

$\rule{90mm}{2pt}$

MORE TO KNOW

$\boxed{\begin{array}{c|c} \bf \underline{ \: \: Complex \: number \: \: } & \bf \underline{ \: \: arg(z) \: \: } \\ & \\ \sf x + iy & \sf {tan}^{ - 1}\bigg |\dfrac{y}{x} \bigg| \\ \\ \sf - x + iy & \sf \pi - {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \\ \\ \sf - x - iy & \sf - \pi + {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \\ \\ \sf x - iy & \sf - {tan}^{ - 1}\bigg |\dfrac{y}{x}\bigg | \end{array}}$

Answer 2

$\pmb{ \sf \boxed{ \bold{ \gray{solution}}}}$

The Modulus of the complex number

z = a + ib i |z| = r = √a² + b²

Where a = r cos ∅ + ir sin∅

Let r cos∅ + ir sin ∅ = -3 + 0i

Therefore x cos ∅ = -3 and r sin ∅ = 0

Where r > 0 and -π < ∅ ≤ π

Now, squaring and adding, we get

=> r² cos² ∅ + r² sin² ∅ = (-3)² + (0)²

=> r² (cos² ∅ + sin² ∅) = (-3)² + (0)²

=> r² (cos²∅ + sin²∅) = 9+0

=> r² (cos² ∅ + sin² ∅) = 9

=> r² = 9

=> r = √9

=> r = 3 [ r > 0 ]

Therefore,

=> r(cos∅) = -3

=> 3(cos∅) = -3

=> cos∅ = -1

And,

=> r(sin∅) = 0

=> 3(sin∅) = -3

=> sin ∅ = 0

Hence, the polar form of -3 is r(cos∅ + sin∅) i.e.,

3(cos∅ + isin∅)