Question

give the polar form of complex number -1-i

Answer 1

for polar form we need to convert it into the form of cos and sin functions
we can do this by dividing it by
√[ (a^2) + (b^2) ]
where a and b are coefficient
so dividing and multiplying by √2
$\sqrt{ { (- 1)}^{2} + {( - 1)}^{2} }$
-(√2) [ (1/√2) + (1/√2)i ]
now we can substitute
-(√2) [cos45° + (sin45°)i ]
while some also write it as
-(√2)cis(45°)

Answer 2

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

Assumption

z = (-1 - i)

It is clear that (-1 - i) lies in 4 Quadrant

z = r(cosθ + isinθ)

Now,

rcosθ = -1

rsinθ = -1

Now here,

$\textbf{\underline{Squarring\;both\;sides :-}}$

r² = 2

r = √2

Hence,

${\boxed{\sf\:{cos\theta=\dfrac{-1}{\sqrt{2}}}}}$

Also,

${\boxed{\sf\:{sin\theta=\dfrac{-1}{\sqrt{2}}}}}$

tanθ = 1

tanα = |tanθ| = 1

${\boxed{\sf\:{\alpha=\dfrac{\pi}{4}}}}$

Then,

$\textbf{\underline{It\;lies\;in\;3\;Quadrant}}$

Hence,

θ = (π - α)

${\boxed{\sf\:{\dfrac{\pi}{4}-\pi =\dfrac{-3\pi}{4}}}}$

Thus,

$\Large{\boxed{\sf\:{z=\sqrt{2}[cos\dfrac{-3 \pi}{4}+isin\dfrac{-3 \pi}{4}}}}$