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kari74 [83]
3 years ago
12

Efectuati calculele:

Mathematics
1 answer:
NNADVOKAT [17]3 years ago
8 0
\sqrt{(5 - 1)^{2}} +  \sqrt{20}

\sqrt{4^{2}} +  \sqrt{4 * 5}

\sqrt{16} +  \sqrt{4} \sqrt{5} 


4 + 2 \sqrt{5} 


4\sqrt{5} + 2 \sqrt{5} 


6 \sqrt{5}


\sqrt{(10 - 3)^{2}} +  \sqrt{20 * 18}

\sqrt{7^{2}}  +  \sqrt{4 * 9 * 5 * 2}

\sqrt{49} +  \sqrt{36 * 10}

7 +  \sqrt{36}\sqrt{10}

7 + 6\sqrt{10}

7\sqrt{10} + 6\sqrt{10}

13\sqrt{10}


\sqrt[3]{(3 + 1)^{2}} -  \sqrt{(3 - 2)x^{2}

\sqrt[3]{4^{2}} -  \sqrt{x^{2}

\sqrt[3]{16} - x[tex] \\[tex] \sqrt[3]{8 * 2} - x

\sqrt[3]{8} \sqrt[3]{2} - x


2\sqrt[3]{2} - x




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Answer:

z=5\left(\cos \left(\dfrac{3\pi}{2}\right)+i\sin \left(\dfrac{3\pi}{2}\right)\right)

Step-by-step explanation:

If a complex number is z=a+ib, then the trigonometric form of complex number is

z=r(\cos \theta +i\sin \theta)

where, r=\sqrt{a^2+b^2} and \tan \theta=\dfrac{b}{a}, \theta is called the argument of z, 0\leq \theta\leq 2\pi.

The given complex number is -5i.

It can be rewritten as

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Here, a=0 and b=-5. \theta lies in 4th quadrant.

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\theta=2\pi -\dfrac{\pi}{2}    [\because \text{In 4th quadrant }\theta=2\pi-\theta]

\theta=\dfrac{3\pi}{2}

So, the trigonometric form is

z=5\left(\cos \left(\dfrac{3\pi}{2}\right)+i\sin \left(\dfrac{3\pi}{2}\right)\right)

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