Express $z = (6 − 2i)(4 − 7i)$ in polar form and calculate $z^2$.

Question

This is a revised version of my previous question, "If $z = (6 − 2i)(4 − 7i)$, find $z^2$".

Question:

Express $z = (6 − 2i)(4 − 7i)$ in polar form and calculate $z^2$. Express the results both in polar and rectangular forms.

Attempt:

$z = 10-50i$
$r= 10{\sqrt 26}$
$\theta=\tan^{-1}\left(\frac{-50}{10}\right)$
$\theta=-1.373400767$
4th Quadrant, so $\theta=2\pi-(-1.373400767)=7.656586074$
$10{\sqrt 26}r[cos(7.656586074) + i sin(7.656586074)]$=z in polar form.

$z^2$ in rectangular form $=$ $(10-50i)(10-50i)$ $=$ $-2400-1000i$
$r=2600$
$\theta=\tan^{-1}\left(\frac{-1000}{-2400}\right)$
$\theta=0.3947911197$
3rd Quadrant, so $\theta=0.3947911197-\pi=-2.746801534$
$z^2$ in polar form $=$ $2600[cos(-2.746801534) + i sin(-2.746801534)]$

Is this correct?

Answer 1

Well, we have:

$$(6-2i)(4-7i)=6\cdot4+6\cdot(-7i)-2i\cdot4-2i\cdot(-7i)=24-42i-8i-14=10-50i\tag1$$

So:

• The magnitude is: $$\left|(6-2i)(4-7i)\right|=\left|10-50i\right|=\sqrt{10^2+50^2}=10\sqrt{26}\tag2$$
• The phase/argument is: $$\arg\left((6-2i)(4-7i)\right)=\arg\left(10-50i\right)=\frac{3\pi}{2}+\arctan\left(\frac{10}{50}\right)=\frac{3\pi}{2}+\arctan\left(\frac{1}{5}\right)\tag3$$

So:

$$(6-2i)(4-7i)=10\sqrt{26}\cdot\left(\cos\left(\frac{3\pi}{2}+\arctan\left(\frac{1}{5}\right)\right)+\sin\left(\frac{3\pi}{2}+\arctan\left(\frac{1}{5}\right)\right)i\right)\tag4$$

And:

$$\left((6-2i)(4-7i)\right)^2=\left(10\sqrt{26}\cdot\exp\left(\left(\frac{3\pi}{2}+\arctan\left(\frac{1}{5}\right)\right)i\right)\right)^2=$$ $$2600\cdot\exp\left(2\cdot\left(\frac{3\pi}{2}+\arctan\left(\frac{1}{5}\right)\right)i\right)=2600\cdot\exp\left(\left(3\pi+2\arctan\left(\frac{1}{5}\right)\right)i\right)=$$ $$2600\cdot\left(\cos\left(3\pi+2\arctan\left(\frac{1}{5}\right)\right)+\sin\left(3\pi+2\arctan\left(\frac{1}{5}\right)\right)i\right)\tag5$$