$\left|a+bi\right|\:=\sqrt{\left(a+bi\right)\left(a-bi\right)}=\sqrt{a^2+b^2}\\\mathrm{With\:}a=4,\:b=2\\=\sqrt{4^2+2^2}\\Refine\\=\sqrt{20}\\\sqrt{20}=2\sqrt{5}\\=2\sqrt{5}$

How to get answer for number 2: | 5-i |

$\left|a+bi\right|\:=\sqrt{\left(a+bi\right)\left(a-bi\right)}=\sqrt{a^2+b^2}\\\mathrm{With\:}a=5,\:b=-1\\=\sqrt{5^2+\left(-1\right)^2}\\=\sqrt{26}$

Number 3 how to get answer: | -3i |

$\left|a+bi\right|\:=\sqrt{\left(a+bi\right)\left(a-bi\right)}=\sqrt{a^2+b^2}\\\mathrm{With\:}a=0,\:b=-3\\=\sqrt{0^2+\left(-3\right)^2}\\Refine\\=\sqrt{9}\\\sqrt{9}\\\mathrm{Factor\:the\:number:\:}\:9=3^2\\=\sqrt{3^2}\\\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a^n}=a\\\sqrt{3^2}=3\\= 3$

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3 years ago

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Please help with three questions!! Worth 98 points!!

qwelly [4]

Answer:

Please show us the questions and will help further this discussion.

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3 years ago

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