From Polar to Cartersian Complex Numbers

**Paymemoney** · November 2nd 2010, 04:39 PM

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

**Archie Meade** · November 2nd 2010, 04:56 PM

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.

**Paymemoney** · November 2nd 2010, 06:01 PM

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

**Soroban** · November 2nd 2010, 08:05 PM

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

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

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

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

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

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

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

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

**Paymemoney** · November 2nd 2010, 10:36 PM

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

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