Question

(z)=-2*i-2 записать в тригонометрической и показательной форме число,если z^10

Answer 1

Ответ:можно точнее?

Пошаговое объяснение:пж

Answer 2

To express the complex number $z = -2i - 2$ in trigonometric and exponential (polar) form, we need to find its magnitude (modulus) and argument (angle).

The modulus of a complex number is given by: $|z| = \sqrt{\text{Re}(z)^2 + \text{Im}(z)^2}$ where $\text{Re}(z)$ is the real part of $z$ and $\text{Im}(z)$ is the imaginary part of $z$.

The argument of a complex number is given by: $\text{arg}(z) = \text{atan2}(\text{Im}(z), \text{Re}(z))$ where $\text{atan2}$ is the arctangent function that takes into account the signs of both the real and imaginary parts to determine the correct quadrant of the complex plane.

Let's calculate the modulus and argument of $z$: $\text{Re}(z) = -2$ $\text{Im}(z) = -2$ $|z| = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$

Now, to find the argument, we'll use the arctangent function: $\text{arg}(z) = \text{atan2}(-2, -2)$ Using a calculator, we get: $\text{arg}(z) \approx -135^\circ$

Now we have the polar form of $z$: $z = 2\sqrt{2} \cdot (\cos(-135^\circ) + i\sin(-135^\circ))$

To find $z^{10}$ in the polar form, we simply raise the magnitude to the power of 10 and multiply the argument by 10: $z^{10} = [2\sqrt{2}]^{10} \cdot \left(\cos(10 \cdot -135^\circ) + i\sin(10 \cdot -135^\circ)\right)$

Let's calculate $z^{10}$ step by step: $[2\sqrt{2}]^{10} = 2^{10} \cdot (2)^{\frac{1}{2} \cdot 10} = 2^{10} \cdot 2^5 = 2^{15} = 32768$

$10 \cdot -135^\circ = -1350^\circ$

Since the angle is given in degrees, we need to convert it to radians to use it in trigonometric functions: $\text{angle in radians} = \frac{\text{angle in degrees} \cdot \pi}{180} = \frac{-1350 \cdot \pi}{180} = -\frac{9}{2}\pi$

Now, we can write $z^{10}$ in the polar (exponential) form: $z^{10} = 32768 \cdot \left(\cos\left(-\frac{9}{2}\pi\right) + i\sin\left(-\frac{9}{2}\pi\right)\right)$