Question

if z=1-i find the imaginary part of z^6

Answer 1

Answer:

$\mathrm{Im}(z^6)=4\sqrt{2}$

Step-by-step explanation:

Firstly, we will convert our complex number z into an exponential form.

Let:

$z=x+iy$

This can be converted into the form:

$z=re^{i\theta}$

$r \:\mathrm{(modulus\:of\:z)}=\sqrt{x^2+y^2}$  

$\displaystyle \theta\:\mathrm{(argument\:of\:z)}=\arctan\left(\frac{y}{x}\right)$

Hence, our complex number z=1-i can be converted into this form by first finding its modulus and argument.

$r=\sqrt{1+(-1)^2}=\sqrt{2}$

$\displaystyle \theta=\arctan\left(\frac{-1}{1}\right)$

$\displaystyle \theta=-\frac{\pi}{4}$

Substitute both the values for r and theta:

$z=re^{i\theta}$

$\displaystyle z=\sqrt{2} e^{i\frac{\pi}{4} }$

Now, we will apply the power on both sides:

$\displaystyle z=\sqrt{2} e^{i\frac{\pi}{4} }$

$\displaystyle z^6=\left(\sqrt{2} e^{i\frac{\pi}{4} }\right)^6$

$z^6=8e^{i\frac{3}{2}\pi}$

Using the Euler's formula, we will now convert our exponential form to the polar form:

$e^{i\theta}=\cos \theta+i\sin \theta$

$\displaystyle e^{i\frac{3}{4}\pi}=-\frac{\sqrt{2} }{2}+i \frac{\sqrt{2} }{2}$

$z^6=8e^{i\frac{3}{2}\pi}$

$\displaystyle z^6=8\left(-\frac{\sqrt{2} }{2}+i \frac{\sqrt{2} }{2}\right)$

$\displaystyle z^6=-4\sqrt{2}+i\:4\sqrt{2}$

Thus, we have our Imaginary part:

$\mathrm{Im}(z^6)=4\sqrt{2}$

Answer 2

Answer:

8

Step-by-step explanation:

simply calculate z6 by calculating (1-i)2 three times. the multiplication will come 8i so 8 is the answer