Thread: From Polar to Cartersian Complex Numbers

1. From Polar to Cartersian Complex Numbers

Can anyone help me on the following question

1) By first converting to polar form evaluate: $\frac{(-1-j)^6}{(1+j)^4}$

P.S

2. Originally Posted by Paymemoney
Can anyone help me on the following question

1) By first converting to polar form evaluate: $\frac{(-1-j)^6}{(1+j)^4}$

P.S
Take the numerator and denominator as a pair of complex numbers.

The modulus of both the numerator and denominator is $\sqrt{2}$ by Pythagoras' theorem.

The angle of the complex number $-1-j$ in the numerator is $\frac{5{\pi}}{4}$

The angle of the complex number $1+j$ in the denominator is $\frac{{\pi}}{4}$

Use De Moivre's theorem to calculate the powers, thereby writing both numerator and denominator in polar form.

To divide 2 complex numbers in polar form, you calculate the ratio of their moduli and subtract their angles.

Then with the result evaluated in polar form, you may convert that to Cartesian Form if necessary.

3. this is what have done, but not the same as the answer

$cis(\frac{30\pi}{4}-{\pi})$

$cis(\frac{26\pi}{4})$

$cos(\frac{26\pi}{4})+jsin(\frac{26\pi}{4})$

book's answer is 2j

4. Hello, Paymemoney!

$\text{1) By first converting to polar form, evaluate: }\;\dfrac{(\text{-}1-j)^6}{(1+j)^4}$

Numerator

$\text{-}1 - j \:=\:\sqrt{2}\left(\cos\frac{5\pi}{4} + i\sin\frac{5\pi}{4}\right)$

$(\text{-}1-j)^6 \;=\;(\sqrt{2})^6 \left(\cos\frac{5\pi}{4} + i\sin\frac{5\pi}{4}\right)^6$

. . . . . . . $=\;8\left(\cos\frac{15\pi}{2} + i\sin\frac{15\pi}{2}\right)$

. . . . . . . $=\;8( 0 -i) \;=\;-8i$

Denominator

$1 + j \:=\:\sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)$

$(1+j)^4 \;=\;(\sqrt{2})^4\left(\cos\frac{\pi}{4} +i\sin\frac{\pi}{4}\right)^4$

. . . . . . . $=\;4(\cos \pi + i \sin\pi)$

. . . . . . . $=\;4(-1 + 0) \;=\;-4$

$\text{Therefore: }\;\dfrac{(\text{-}1 - j)^6}{(1+j)^4} \;=\;\dfrac{-8j}{-4} \;=\;2j$

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It would have been easier without the conversion . . .

The numerator has: . $\text{-}1 - j \:=\:-1(1+j)$

. . Hence: . $\bigg[\text{-}1(1+j)\bigg]^6 \;=\;(\text{-}1)^6 (1+j)^6 \;=\;(1+j)^6$

The fraction becomes: . $\dfrac{(1+j)^6}{(1+j)^4} \;=\;(1+j)^2 \;=\;1 + 2j + i^2 \;=\; 2i$

5. oh i know where i went wrong, silly arithmetic error.