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

Write the following in the polar form- (i)−1−i - 1 - {\text{i}} (ii)1−i1 - {\text{i}}

Explanation

Solution

The polar form of a + ib{\text{a + ib}} is rcosθ + irsinθ{\text{rcos}}\theta {\text{ + irsin}}\theta where r = a2+b2{\text{r = }}\sqrt {{{\text{a}}^2} + {{\text{b}}^2}} , cos⁡θ=ar\cos \theta = \dfrac{{\text{a}}}{{\text{r}}} and sin⁡θ=br\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−i - 1 - {\text{i}} which is in the form of a + ib{\text{a + ib}} where a=−1 - 1 and b= −1 - 1
We have to write it in polar form which is z=rcosθ + irsinθ{\text{rcos}}\theta {\text{ + irsin}}\theta wherer = a2+b2{\text{r = }}\sqrt {{{\text{a}}^2} + {{\text{b}}^2}} , cos⁡θ=ar\cos \theta = \dfrac{{\text{a}}}{{\text{r}}} and sin⁡θ=br\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 r = a2+b2{\text{r = }}\sqrt {{{\text{a}}^2} + {{\text{b}}^2}} .
⇒\Rightarrow r=(−1)2+(−1)2\sqrt {{{\left( { - 1} \right)}^2} + {{\left( { - 1} \right)}^2}}
On simplifying we get,
⇒\Rightarrow r=1+1=2\sqrt {1 + 1} = \sqrt 2
Now that we know the value of r we can find the values of θ{{\theta }}.
⇒cos⁡θ=ar=−12\Rightarrow \cos \theta = \dfrac{{\text{a}}}{{\text{r}}} = \dfrac{{ - 1}}{{\sqrt 2 }} And since we know that cos⁡3π4=−12\cos \dfrac{{3\pi }}{4} = \dfrac{{ - 1}}{{\sqrt 2 }}
So we can find the value of θ\theta
⇒cos⁡θ=cos⁡3π4⇒θ=3π4\Rightarrow \cos \theta = \cos \dfrac{{3\pi }}{4} \Rightarrow \theta = \dfrac{{3\pi }}{4}
On putting the values or r and θ{{\theta }}in the value of z, we get the value of z in polar form
z=2(cos⁡3π4+sin⁡3π4) = \sqrt 2 \left( {\cos \dfrac{{3\pi }}{4} + \sin \dfrac{{3\pi }}{4}} \right)
(ii) Given complex number z=1−i1 - {\text{i}} which is in the form of a + ib{\text{a + ib}} where a= 11 and b=−1 - 1
We have to write it in polar form which is z=rcosθ + irsinθ{\text{rcos}}\theta {\text{ + irsin}}\theta where r = a2+b2{\text{r = }}\sqrt {{{\text{a}}^2} + {{\text{b}}^2}} , cos⁡θ=ar\cos \theta = \dfrac{{\text{a}}}{{\text{r}}} and sin⁡θ=br\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 r = a2+b2{\text{r = }}\sqrt {{{\text{a}}^2} + {{\text{b}}^2}}
⇒\Rightarrow r=(1)2+(−1)2\sqrt {{{\left( 1 \right)}^2} + {{\left( { - 1} \right)}^2}}
On simplifying we get,
⇒\Rightarrow r=1+1=2\sqrt {1 + 1} = \sqrt 2
Now that we know the value of r we can find the value of θ{{\theta }}.
⇒cos⁡θ=ar=12\Rightarrow \cos \theta = \dfrac{{\text{a}}}{{\text{r}}} = \dfrac{1}{{\sqrt 2 }} =cos⁡π4 = \cos \dfrac{\pi }{4} [ as cos⁡π4=12\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }} ]
⇒θ=π4\Rightarrow \theta = \dfrac{\pi }{4}
On putting the values or r and θ{{\theta }}in the value of z, we get
z=2(cos⁡π4+sin⁡π4) = \sqrt 2 \left( {\cos \dfrac{\pi }{4} + \sin \dfrac{\pi }{4}} \right).

Note: The polar form of a + ib{\text{a + ib}} can also be written as (r,θ)\left( {{\text{r,}}\theta } \right).So the polar form of −1−i - 1 - {\text{i}} can be written as (2,3π4)\left( {\sqrt 2 ,\dfrac{{3\pi }}{4}} \right) and the polar form of 1−i1 - {\text{i}} can be written as (2,π4)\left( {\sqrt 2 ,\dfrac{\pi }{4}} \right) .In the complex number a + ib{\text{a + ib}}, a{\text{a}} is the real part and b{\text{b}} is the imaginary part of the complex number.