Question

Write the following in the polar form- (i)$ - 1 - {\text{i}}$ (ii)$1 - {\text{i}}$

Answer 1

Hint: The polar form of ${\text{a + ib}}$ is ${\text{rcos}}\theta {\text{ + irsin}}\theta $ where ${\text{r = }}\sqrt {{{\text{a}}^2} + {{\text{b}}^2}} $, $\cos \theta = \dfrac{{\text{a}}}{{\text{r}}}$ and $\sin \theta = \dfrac{{\text{b}}}{{\text{r}}}$ . We can use these formulas to find the polar form of the given complex numbers.

Complete step-by-step answer:
(i)Given complex number z=$ - 1 - {\text{i}}$ which is in the form of ${\text{a + ib}}$ where a=$ - 1$ and b= $ - 1$
We have to write it in polar form which is z=${\text{rcos}}\theta {\text{ + irsin}}\theta $ where${\text{r = }}\sqrt {{{\text{a}}^2} + {{\text{b}}^2}} $, $\cos \theta = \dfrac{{\text{a}}}{{\text{r}}}$ and $\sin \theta = \dfrac{{\text{b}}}{{\text{r}}}$.
Since we know the values of ‘a’ and ‘b’, so put the values of ‘a’ and ‘b’ in ${\text{r = }}\sqrt {{{\text{a}}^2} + {{\text{b}}^2}} $ .
$ \Rightarrow $ r=$\sqrt {{{\left( { - 1} \right)}^2} + {{\left( { - 1} \right)}^2}} $
On simplifying we get,
$ \Rightarrow $ r=$\sqrt {1 + 1} = \sqrt 2 $
Now that we know the value of r we can find the values of ${{\theta }}$.
$ \Rightarrow \cos \theta = \dfrac{{\text{a}}}{{\text{r}}} = \dfrac{{ - 1}}{{\sqrt 2 }}$ And since we know that $\cos \dfrac{{3\pi }}{4} = \dfrac{{ - 1}}{{\sqrt 2 }}$
So we can find the value of $\theta $
$ \Rightarrow \cos \theta = \cos \dfrac{{3\pi }}{4} \Rightarrow \theta = \dfrac{{3\pi }}{4}$
On putting the values or r and ${{\theta }}$in the value of z, we get the value of z in polar form
z$ = \sqrt 2 \left( {\cos \dfrac{{3\pi }}{4} + \sin \dfrac{{3\pi }}{4}} \right)$
(ii) Given complex number z=$1 - {\text{i}}$ which is in the form of ${\text{a + ib}}$ where a= $1$ and b=$ - 1$
We have to write it in polar form which is z=${\text{rcos}}\theta {\text{ + irsin}}\theta $ where ${\text{r = }}\sqrt {{{\text{a}}^2} + {{\text{b}}^2}} $, $\cos \theta = \dfrac{{\text{a}}}{{\text{r}}}$ and $\sin \theta = \dfrac{{\text{b}}}{{\text{r}}}$.
Since we know the values of a and b so put the values of a and b in ${\text{r = }}\sqrt {{{\text{a}}^2} + {{\text{b}}^2}} $
$ \Rightarrow $ r=$\sqrt {{{\left( 1 \right)}^2} + {{\left( { - 1} \right)}^2}} $
On simplifying we get,
$ \Rightarrow $ r=$\sqrt {1 + 1} = \sqrt 2 $
Now that we know the value of r we can find the value of ${{\theta }}$.
$ \Rightarrow \cos \theta = \dfrac{{\text{a}}}{{\text{r}}} = \dfrac{1}{{\sqrt 2 }}$ $ = \cos \dfrac{\pi }{4}$ [ as $\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}$ ]
$ \Rightarrow \theta = \dfrac{\pi }{4}$
On putting the values or r and ${{\theta }}$in the value of z, we get
z$ = \sqrt 2 \left( {\cos \dfrac{\pi }{4} + \sin \dfrac{\pi }{4}} \right)$.

Note: The polar form of ${\text{a + ib}}$ can also be written as $\left( {{\text{r,}}\theta } \right)$.So the polar form of $ - 1 - {\text{i}}$ can be written as $\left( {\sqrt 2 ,\dfrac{{3\pi }}{4}} \right)$ and the polar form of $1 - {\text{i}}$ can be written as $\left( {\sqrt 2 ,\dfrac{\pi }{4}} \right)$ .In the complex number ${\text{a + ib}}$, ${\text{a}}$ is the real part and ${\text{b}}$ is the imaginary part of the complex number.