Answer:
The complex number which has a distance of √17 from the origin to the complex plane is:
Option: D
\(4-i\)
Step-by-step explanation:
We know that the distance of a complex number of the form a+ib from the origin ( (0,0) or 0+0.i ) is given by:
\(\sqrt{a^2+b^2}\)
Hence, we will calculate this expression in each of the options and check which is equal to √17
A)
\(2+15\cdot i\)
Here a=2 and b=15
Hence,
\(\sqrt{a^2+b^2}=\sqrt{2^2+(15)^2}\\\\\sqrt{a^2+b^2}=\sqrt{4+225}\\\\\\\sqrt{a^2+b^2}=\sqrt{229}
eq \sqrt{17}\)
Option: A is incorrect.
B)
\(17+i\)
Here a=17 and b=1
Hence,
\(\sqrt{a^2+b^2}=\sqrt{(17)^2+1^2}\\\\\sqrt{a^2+b^2}=\sqrt{290}
eq \sqrt{17}\)
Hence, option: B is incorrect.
C)
\(20-3i\)
Here a=20 and b= -3
\(\sqrt{a^2+b^2}=\sqrt{(20)^2+(-3)^2}\\\\\sqrt{a^2+b^2}=\sqrt{409}
eq \sqrt{17}\)
Hence, option: C is incorrect.
D)
\(4-i\)
Here a=4 and b= -1
\(\sqrt{a^2+b^2}=\sqrt{(4)^2+(-1)^2}\\\\\sqrt{a^2+b^2}=\sqrt{17}\)
Hence, option: D is correct.
