How do you convert #2 - 2i# to polar form?

Question

Madelyn Adams · Accepted Answer

#(2,-2)to(2sqrt2,-pi/4)#

To convert from #color(blue)"cartesian to polar form"#

That is #(x,y)to(r,theta)#

#color(orange)"Reminder" color(red)(|bar(ul(color(white)(a/a)color(black)(r=sqrt(x^2+y^2))color(white)(a/a)|)))#

and #color(red)(|bar(ul(color(white)(a/a)color(black)(theta=tan^-1(y/x))color(white)(a/a)|)))#

here x = 2 and y = -2

#rArrr=sqrt(2^2+(-2)^2)=sqrt8=2sqrt2#

Now 2 - 2i is in the 4th quadrant, hence we must ensure that #theta# is in the 4th quadrant.

#theta=tan^-1((-2)/2)=tan^-1(-1)=-pi/4" in 4th quadrant"#

#rArr2-2i=(2,-2)to(2sqrt2,-pi/4)#

James Adkins · Answer

undefined To convert $2 - 2i$ to polar form, follow these steps:

1. Find the magnitude (or absolute value) of the complex number using the formula $r = \sqrt{a^2 + b^2}$, where $a$ is the real part and $b$ is the imaginary part.
2. Find the angle (or argument) of the complex number using the formula $\theta = \arctan{\left(\frac{b}{a}\right)}$, where $a$ and $b$ are the real and imaginary parts, respectively.
3. Express the complex number in polar form as $r(\cos{\theta} + i\sin{\theta})$.

Given $2 - 2i$, where the real part $a = 2$ and the imaginary part $b = -2$, we can proceed as follows:

1. Calculate the magnitude:
$$ r = \sqrt{(2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} $$

2. Calculate the angle:
$$ \theta = \arctan{\left(\frac{-2}{2}\right)} = \arctan{(-1)} $$

3. Determine the angle in the appropriate quadrant. Since both the real and imaginary parts are negative, the angle lies in the third quadrant, so we need to add $\pi$ to the result of the arctan function. Thus:
$$ \theta = \arctan{(-1)} + \pi = -\frac{\pi}{4} + \pi = \frac{3\pi}{4} $$

4. Express the complex number in polar form:
$$ 2 - 2i = 2\sqrt{2} \left(\cos{\frac{3\pi}{4}} + i\sin{\frac{3\pi}{4}}\right) $$

So, the polar form of $2 - 2i$ is $2\sqrt{2}\left(\cos{\frac{3\pi}{4}} + i\sin{\frac{3\pi}{4}}\right)$.