Respuesta :
The distance of a complex number a+bi from the origin on the complex plane is given by the formula:
[tex]\sqrt{a^2+b^2}[/tex]
a.
[tex]2+15i \\ a=2 \\ b=15 \\ \Downarrow \\ d=\sqrt{2^2+15^2}=\sqrt{4+225}=\sqrt{229}[/tex]
b.
[tex]17+i \\ a=17 \\ b=1 \\ \Downarrow \\ d=\sqrt{17^2+1^2}=\sqrt{289+1}=\sqrt{290}[/tex]
c.
[tex]20-3i \\ a=20 \\ b=-3 \\ \Downarrow \\ d=\sqrt{20^2+(-3)^2}=\sqrt{400+9}=\sqrt{409}[/tex]
d.
[tex]4-i \\ a=4 \\ b=-1 \\ \Downarrow \\ d=\sqrt{4^2+(-1)^2}=\sqrt{16+1}=\sqrt{17}[/tex]
The answer is D.
[tex]\sqrt{a^2+b^2}[/tex]
a.
[tex]2+15i \\ a=2 \\ b=15 \\ \Downarrow \\ d=\sqrt{2^2+15^2}=\sqrt{4+225}=\sqrt{229}[/tex]
b.
[tex]17+i \\ a=17 \\ b=1 \\ \Downarrow \\ d=\sqrt{17^2+1^2}=\sqrt{289+1}=\sqrt{290}[/tex]
c.
[tex]20-3i \\ a=20 \\ b=-3 \\ \Downarrow \\ d=\sqrt{20^2+(-3)^2}=\sqrt{400+9}=\sqrt{409}[/tex]
d.
[tex]4-i \\ a=4 \\ b=-1 \\ \Downarrow \\ d=\sqrt{4^2+(-1)^2}=\sqrt{16+1}=\sqrt{17}[/tex]
The answer is D.