Question

Convert the given complex number into polar form : 1 - i


No spam!! ​

Answer 1

Answer:

Given, z=1−i

Let rcosθ=1andrsinθ=−1

On squaring and adding, we obtain

r

2

cos

2

θ+r

2

sin

2

θ=1

2

+(−1)

2

⇒r

2

(cos

2

θ+sin

2

θ)=2

⇒r

2

=2

⇒r=

2

(since,r>0 )

∴

2

cosθ=1 and

2

sinθ=−1

∴θ=−

4

π

(As θ lies in fourth quadrant.)

So, the polar form is

∴1−i=rcosθ+irsinθ=

2

cos(

4

−π

)+i

2

sin(

4

−π

)

=

2

[cos(

4

−π

)+isin(

4

−π

)]

Answer 2

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

Given complex number is

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

Let assume that

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

can be further rewritten as

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

On comparing with Real and Imaginary parts, we get

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

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

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

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

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

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

$\bf\implies \:r = \sqrt{2}$

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

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

and

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

$\rm \implies\:\theta \: lies \: in \: {4}^{th} \: quadrant$

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

So, on substituting the values in equation (1), we get

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

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

More to know

Argument of complex number

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