Question

Convert in polar form   -1-i

Answer 1

$\huge{HEY MATE!!! }$

HERE IS UR ANSWER ⏬⏬⏬⏬⏬

❇➡ Let z = -1-i. Then,

               r = l z l = √ (-1)2 + (-1)2 = √2

 

Let tan α = | Im (z) / Re (z) |. Then,

      tan α = | -1/ -1 | = 1 or α = π/4

Since the point ( -1, -1) representing z lies in the third quadrant. Therefore, the argument of z is given by 

                   θ = - (π - α) = - ( π - π/4) = -3π/4

So, the polar form of z = -1-i is

                  z = r ( cosθ + i sin θ )

                     = √2{ cos (-3π/4) + i sin (-3π/4) }

                     = √2 ( cos 3π/4 - i sin 3π/ 4) is the required answer

Hope it helps.

Cheer!!!!!!!!!!!!!!!!!

Answer 2

$\huge{Hello Friend}$

The answer of u r question is..✌️✌️

✨Convert in polar form -1-i

Ans:✍️✍️✍️✍️✍️✍️

Let,

$r \: = z \: = \sqrt{( - 1)} 2 + ( - 1)2 = \sqrt{2}$



Let,


$tan \: a \: = | \frac{m(2)}{r \: e \: (2)} |$


Then,

$tan \: a \: = | \frac{ - 1}{ - 1} | = 1 = \frac{\pi}{4}$


$= tita = - (\pi - a) = - (\pi - \frac{\pi}{4} ) = - 3 \frac{\pi}{4}$

so,

Polar form of -1-i is....

$z = n(cos \: 0 + i \: sin \: 0)$


$= \sqrt{2} (cos \: - 3 \frac{\pi}{4} ) + i \: sin \: ( - 3 \frac{\pi}{4} )$

$= \sqrt{2} (\: {cos3\pi}{4 - i \: sin \: \frac{3\pi}{4} } )$

Thank you..⭐️⭐️