Question

convert into polar form 1+ i root 3​

Answer 1

Answer:

Z=1+i

3

=2(

2

1

+i

2

3

)...........(1)

Let Polar form of Given Equation be Z=rcosθ+ir sinθ...............(2)

Comparing (1) and (2)

we can write,

rcosθ=2∗

2

1

=2cos60

o

rsinθ=2∗

2

1

=2sin60

o

∴Z=2(cos60

o

+isin60

o

)

Answer 2

Solution :-

$\bf Polar \ form \ : \ z = r cos \theta+ i \ r sin \theta$

Here,

$\sf z = 1+i\sqrt{3}$

$\longrightarrow\sf rcos \theta = 1 \\\\\longrightarrow r^2 cos^2 \theta= 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ .................(1)$

$\longrightarrow \sf r sin \theta = \sqrt{3} \\\\\longrightarrow r^2 sin^2 \theta= 3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ .................(2)$

Adding eq (2) and eq (1),

$\longrightarrow \sf r^2 cos \theta + r^2 sin \theta = 1+ 3\\\\\longrightarrow r^2(cos \theta+sin \theta) = 4 \\\\\longrightarrow r^2 \times 1 = 4 \\\\\longrightarrow \bf r = 2$

Modulus = 2

$\hookrightarrow \sf 2 cos \theta = 1 \\\\\hookrightarrow\bf cos \theta = \dfrac{1}{2}$

$\hookrightarrow \sf 2 sin \theta = \sqrt{3} \\\\\hookrightarrow \bf sin \theta= \dfrac{\sqrt{3}} {2}$

$\sf sin \theta$ and $\sf cos \theta$ are positive.

So, the argument lies in the first quadrant.

$\sf sin \theta = \dfrac{\sqrt{3}}{2}$ and $\sf cos \theta = \dfrac{1}{2}$ .

$\therefore \sf \theta = 60 ^{\circ}$

Argument = $\sf 60^{\circ}$

                 $= \sf 60\times \dfrac{\pi}{180}\\\\= \dfrac{\pi}{3}$

$\bf Polar \ form = 2 \left( cos \dfrac{\pi}{3}+i \ sin \dfrac{\pi}{3} \right)$