Question

convert into polar from 2+3i/3-2i

Answer 1

To convert from cartesian to polar form

That is (x,y)→(r,θ)

Reminder ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯∣∣∣aar=√x2+y2aa∣∣∣−−−−−−−−−−−−−−−−−

and ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯∣∣aaθ=tan−1(yx)aa∣∣−−−−−−−−−−−−−−−−−−

here x = 3 and y = 2

⇒r=√32+22=√13

Now, 3 + 2i, is in the 1st quadrant so we must ensure that θ is in the 1st quadrant.

θ=tan−1(23)=0.588 radians← in 1st quadrant

Thus (3,2)→(√13,0.588)→(√13,33.69∘)

Answer 2

Let,
$\frac{2 + 3i}{3 - 2i} = r( \cos\theta + i \sin \theta) \\ = > \frac{(2 + 3i)(3 + 2i)}{(3 - 2i)(3 + 2i)} = r( \cos\theta + i \sin\theta ) \\ = > \frac{6 + 4i + 9i - 6}{ {3}^{2} - {(2i)}^{2} } = r( \cos \theta + i \sin \theta ) \\ = > \frac{11i}{11} = r( \cos \theta + i \sin \theta ) \\ = > i = r( \cos \theta + i \sin \theta ) \\ = > 0 + i \times 1 = r \cos \theta + i r\sin \theta$
hence,
$0 = r \cos \theta$
and
$1 = r \sin\theta$
now squaring and adding the equations, we get
${0}^{2} + {1}^{2} = {(r \cos\theta)}^{2} + {(r \sin \theta)}^{2} \\ = > 1 = {r}^{2} { \cos \theta }^{2} + {r}^{2} { \sin\theta }^{2} \\ = > 1 = {r}^{2} ( { \cos \theta }^{2} + { \sin \theta }^{2} ) \\ = > 1 = {r}^{2} \\ = > {r}^{2} = 1 \\ = > r = 1$
Putting the value of r in the above equations, we get
$\cos \theta = 0$
and
$\sin \theta = 1$
hence, the polar form is
$1 \times ( \cos \theta + i \sin \theta ) \\ = \cos \theta + i \sin \theta$