Question

What is (-2√3-2i)^4 equivalent to?
(-2√3-2i)^4 =___ cis ___ or ____
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Answer 1

`\left(-2√(3)-2i\right)^4=\boxed{256\:\text{cis}\left((2\pi)/(3)\right)\right)}\;\;\textsf{or}\;\;\boxed{-128 + 128 √(3)\: i}`

Explanation:

Rectangular form

Rectangular form, also known as the Cartesian form, is a way to represent a complex number using its real and imaginary parts.

A complex number is represented in the form z = a + bi, where a is the real part and b is the imaginary part.

To express (-2√3 - 2i)⁴ in rectangular form, expand using the binomial theorem.

`\boxed{\begin{minipage}{11cm}\underline{Binomial Theorem}\\\\$\displaystyle (a+b)^n=a^n+\binom{n}{1}a^(n-1)b^1+\binom{n}{2}a^(n-2)b^2+...+\binom{n}{r}a^(n-r)b^r+...+b^n$\\\\\\where $\displaystyle \binom{n}{r} \: = \:^(n)\text{C}_(r) = (n!)/(r!(n-r)!)$\\\end{minipage}}`

`\textsf{If\;\;$(-2√(3)-2i)^4=(a+b)^n$,\;\;then:}`

• 
  `a = -2√(3)`
• 
  `b = -2i`
• 
  `n = 4`

Substituting these values into the binomial theorem formula:

`\displaystyle =(-2√(3))^4+\binom{4}{1}(-2√(3))^(3)(-2i)+\binom{4}{2}(-2√(3))^(2)(-2i)^2+\binom{4}{3}(-2√(3))(-2i)^3+(-2i)^4`

`=144+4(-24√(3))(-2i)+6(12)(-4)+4(-2√(3))(8i)+16`

`=144+192√(3)\:i-288-64√(3)\:i+16`

`=-128+128√(3)\:i`

Therefore, (-2√3 - 2i)⁴ in rectangular form is:

`\boxed{-128 + 128√(3)\:i}`

`\hrulefill`

Polar form

Polar form is a way to represent a complex number using its magnitude (modulus) and argument (angle).

It is expressed in the form r(cos(θ) + i sin(θ)) where r represents the distance from the origin (magnitude) and θ represents the angle with the positive real axis (argument).

`\textsf{The polar form of a complex number\;\;$z=a+bi$\;\;is\;\;$z=r(\cos \theta+i \sin \theta)$}.`

`\textsfThe magnitude of the complex number\;\;$z=a+bi$\;\;is written as $r=`

`\textsf{Using the rectangular form found in the previous part,\;\;$z=-128+128√(3)\:i$.}\\\\\textsf{Therefore:}`

`\begin{aligned}r=|z|&=\sqrt{(-128)^2+(128√(3))^2}\\&=√((16384+49152)\\&=√(65536)\\&=256\end{aligned}`

`\textsf{For a complex number\;\;$z=a+bi$,\;\;the argument, $\theta$, satisfies\;\;$\tan \theta=(b)/(a)$.}`

`\textsf{$\theta=\tan^(-1)\left((b)/(a)\right)$\;\;for\;\;$a > 0$\;\;and\;\;$\theta=\tan^(-1)\left((b)/(a)\right)+ \pi$\;\;for\;\;$a < 0$}\:.`

`\textsf{Since\;\;$a = -128 < 0$,\;\;use the formula $\theta = \tan^(-1)\left((b)/(a)\right)+ \pi$\::}`

`\theta=\tan ^(-1)\left((128√(3))/(-128)\right)+\pi`

`\theta=\tan^(-1)(-√(3))+\pi`

`\theta=-(\pi)/(3)+\pi`

`\theta=(2\pi)/(3)`

Therefore, the polar form of (-128 + 128√3i) is:

`\boxed{256\left (\cos \left((2\pi)/(3)\right)+i \sin\left((2\pi)/(3)\right)\right)}=\boxed{256\:\text{cis}\left((2\pi)/(3)\right)\right)}`