Use de Moivre’s theorem to find the value of the real part of $(-7+24i)^8$

Question

I’ve solved the first part of the question:

Find the modulus and argument of $(-7+24i)$ giving your value of the argument in radian correct to 3 significant figures.

Modulus of $(-7+24i)= \sqrt{(-7)^2+{24}^2} =25$

Arg$(-7+24i)=\tan^{-1}\big(\displaystyle\frac{24}{-7}\big) =1.85$rad

I don’t know how to solve the second part of the question:

Use de Moivre’s theorem to find the value of the real part of $(-7+24i)^8$, correct to 3 significant figures.

How do I change $(-7+24i)^8$ into a form in which I can apply de Moivre’s theorem?

Answer 1

We have $\sqrt{7^2+24^2} = 25$

So, $$-7+24i = 25\left(\frac{-7}{25} + \frac{24}{25}i\right)$$

Let $\frac{-7}{25} = \cos \theta $ and $\frac{24}{25} = \sin\theta$

Now use the fact that $$(\cos\theta+i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$$

Here $n=8$

Answer 2

$$z=-7+24i=25\left(-\frac 7{25}+i\frac{24}{25}\right)$$ So $$\cos\theta=-\frac{7}{25}$$and$$\sin\theta=\frac{24}{25}$$ Now $$z^8=25^8(\cos 8\theta + i\sin 8\theta)$$ All you need to do is write first $\sin 2\theta$ and $\cos2\theta$ in terms of $\sin\theta$ and $\cos\theta$, then $\sin 4\theta$ and $\cos4\theta$, and finally $\sin 8\theta$ and $\cos8\theta$

Answer 3

You know that $-7 + 24i = 25e^{i\phi}$ for an appropriate phase $\phi$, such as the one you found. Therefore, the real part of its eighth power is just $$\text{Re } 25^8 e^{i8\phi} = 25^8 \cos(8\phi)$$