Question

convert the complex number -1+i in thepolar form math +1​

Answer 1

$\large\underline{\sf{Solution-}}$

Given complex number is

$\rm :\longmapsto\: - 1 + i$

Let assume that

$\rm :\longmapsto\: - 1 + i = r(cos\theta + i \: sin\theta) - - - (1)$

can be rewritten as

$\rm :\longmapsto\: - 1 + i = r \: cos\theta + i \:r \: sin\theta$

On comparing Real and Imaginary parts, we get

$\rm :\longmapsto\:rcos\theta = - 1 - - - (2)$

and

$\rm :\longmapsto\:rsin\theta = 1 - - - (3)$

On squaring equation (2) and (3) and adding, we get

$\rm :\longmapsto\: {r}^{2} {cos}^{2}\theta + {r}^{2} {sin}^{2}\theta = {( - 1)}^{2} + {1}^{2}$

$\rm :\longmapsto\: {r}^{2} ({cos}^{2}\theta + {sin}^{2}\theta )= 1 + 1$

$\rm :\longmapsto\: {r}^{2} = 2$

$\bf\implies \:r \: = \: \sqrt{2} - - - (4)$

On substituting the value of r in equation (2) and (3), we get

$\rm :\longmapsto\:cos\theta = - \dfrac{1}{ \sqrt{2} } \: and \: sin\theta = \dfrac{1}{ \sqrt{2} }$

$\bf\implies \:\theta = \pi - \dfrac{\pi}{4} = \dfrac{3\pi}{4}$

So, Equation (1) can be rewritten as

$\rm \implies\:\boxed{ \tt{ \: - 1 + i = \sqrt{2}\bigg(cos\dfrac{3\pi}{4} + i \: sin \dfrac{3\pi}{4} \bigg) \: }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

More to Know

Short cut trick to find the Argument of a complex number z = x + iy

$\begin{gathered}\boxed{\begin{array}{c|c} \bf complex \: no & \bf arg(z) \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf x + iy & \sf {tan}^{ - 1} \bigg|\dfrac{y}{x} \bigg| \\ \\ \sf - x + iy & \sf \pi - {tan}^{ - 1} \bigg|\dfrac{y}{x} \bigg| \\ \\ \sf - x - iy & \sf - \pi + {tan}^{ - 1} \bigg|\dfrac{y}{x} \bigg|\\ \\ \sf x - iy & \sf - {tan}^{ - 1} \bigg|\dfrac{y}{x} \bigg| \end{array}} \\ \end{gathered}$