Question

Convert the complex number from rectangular form to polar form. Show all work for full credit. z=-14+8i

Answer 1

Ai is dumb i swear

Yes, the complex number -14 + 8i can definitely be represented in polar form. To convert the complex number from rectangular form (a + bi) to polar form (r∠θ), where r is the magnitude and θ is the angle:

First, we calculate the magnitude (r) using the formula: r = √(a^2 + b^2)

In this case: a = -14 b = 8

r = √((-14)^2 + (8)^2) r = √(196 + 64) r = √260 r = 2√65

Next, we calculate the angle (θ) using the formula: θ = tan^(-1)(b/a)

In this case: θ = tan^(-1)(8 / -14) θ ≈ -0.5191 radians

Therefore, the polar form of the complex number -14 + 8i is: z = 2√65∠-0.5191

Answer 2

z = 2√65∠-0.5191

Answer 3

So you are given z, which is depicted as:
\[\Large z=x+yi\]

And you need to get it into this form:
\[\Large z=r(cos\theta+isin\theta)\]

In order to do that, you're gonna need to find r, and r is:
\[\Large r = |z|= \sqrt{x^2+y^2}\]

Plug in the values of x and y

Answer 4

plug in

Answer 5

this is easy, you plug in the numbers to see if the equation is right.
🤦🏾‍♀️

Answer 6

BRUH

Answer 7

righ

Answer 8

WRONG

Answer 9

\[z=-14+8\iota\]
compare with \[z=r(cos \theta+\iota sin \theta)\]
\[r cos \theta=-14\]
\[r sin \theta=8\]
square and add
\[r^2(cos^2 \theta+sin^2 \theta=(-14)^2+8^2=196+64=260=4 \times 65\]
\[r=2 \sqrt{65}\]
again divide
\[tan \theta=-8/14=-4/7\]
\[\theta=\tan ^{-1}(-4/7)=-29.74488 \approx -30 ^\circ\]
sin is positive cos is negative so theta lies in second quadrant.
so \theta=180-30=150
\[z=2 \sqrt{65}(150^\circ)\]

Answer 10

\[z=2\sqrt{65}e ^{150 ^\circ~\iota}\]

Answer 11

\[z=2\sqrt{65}(\cos 150+\iota \sin 150)\]