Question

convert the complex number z= -16/(1+i)³ into polar form

Answer 1

For a complex number
$z=a+ib$

The Polar form is given by:

$z = r(\cos \theta + i\sin \theta) \\ \\ where \\ \\ r = \sqrt{a^2+b^2} \\ \\ \theta = arg(z)$

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Here, our complex number is:

$z = -\frac{16}{(1+i)^3}$

We need to bring it into the standard form first.

We will use the identity:

$\boxed{(x+y)^3=x^3+y^3+3xy(x+y)}$

We have:

$z = -\frac{16}{(1+i)^3}\\ \\ \implies z=-\frac{16}{1^3+i^3+3(1)(i)(1+i)}\\ \\ \implies z=-\frac{16}{1-i+3i+3i^2}\\ \\ \implies z=-\frac{16}{1+2i-3}\\ \\ \implies z=-\frac{16}{-2+2i}\\ \\ \implies z=\frac{-16}{-2(1-i)} \\ \\ \implies z = \frac{8}{1-i} \\ \\ \text{Multiplying numerator and denominator by (1+i)} \\ \\ \implies z = \frac{8}{1-i} \times \frac{1+i}{1+i} \\ \\ \implies z = \frac{8(1+i)}{1^2-i^2} \\ \\ \implies z = \frac{8(1+i)}{2} \\ \\ \implies z = 4(1+i) \\ \\ \implies z = 4+4i$

Now, we need magnitude and argument.

Here, comparing it with standard form $z=a+ib$ both a and b are positive. So our complex number lies in the first quadrant. 

The argument of a complex number in the first quadrant is:

$\theta = arg(z) = \tan^{-1} \frac{b}{a} \\ \\ \implies \theta = \tan^{-1} \left( \frac{4}{4} \right) \\ \\ \implies \theta = \tan^{-1} 1 \\ \\ \implies \theta = \frac{\pi}{4}$

Also, the magnitude is:

$r=\sqrt{a^2+b^2} \\ \\ \implies r = \sqrt{4^2+4^2} \\ \\ \implies r = 4\sqrt{2}$


The polar form is:

$z=r(\cos \theta + i\sin \theta) \\ \\ \\ \implies \boxed{z=4\sqrt{2}\left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right)}$