Find the magnitude (or absolute value) of the following complex number.
z=3-2i
Let z be a complex number such that:
z=a+ib
Then the magnitude of z is:
\left| z \right|=\sqrt{a^2+b^2}
Therefore, the magnitude of 3-2i equals:
\sqrt{3 ^ 2 + \left(- 2\right) ^ {2}}=\sqrt{9+4}=\sqrt{13}
\left| 3-2i \right|=\sqrt{13}
z = 3+4i
Let z be a complex number such that:
z=a+ib
Then the magnitude of z is:
\left| z \right|=\sqrt{a^2+b^2}
Therefore, the magnitude of 3+4i equals:
\sqrt {3 ^ 2 + 4 ^ {2}} = \sqrt {9 + 16} = \sqrt {25}=5
\left | 3+4i \right | = 5
z=1+7i
Let z be a complex number such that:
z=a+ib
Then the magnitude of z is:
\left| z \right|=\sqrt{a^2+b^2}
Therefore, the magnitude of 1 + 7i equals:
\sqrt {1 ^ 2 + 7 ^ {2}} = \sqrt {1 + 49} = \sqrt {50}=\sqrt{25} \cdot \sqrt{2}=5\sqrt{2}
\left | 1 + 7i \right | = 5\sqrt {2}
z=-3-6i
Let z be a complex number such that:
z=a+ib
Then the magnitude of z is:
\left| z \right|=\sqrt{a^2+b^2}
Therefore, the magnitude of -3 -6i equals:
\sqrt {\left(-3\right) ^ 2 + \left(-6\right) ^ {2}} = \sqrt {9 + 36} = \sqrt {45}=\sqrt{9} \cdot \sqrt{5}=3\sqrt{5}
\left | -3 -6i \right | = 3 \sqrt {5}
z=11i
Let z be a complex number such that:
z=a+ib
Then the magnitude of z is:
\left| z \right|=\sqrt{a^2+b^2}
Therefore, the magnitude of 7i equals:
\sqrt {0^ 2 + 7^ {2}} = \sqrt {49} =7
\left | 7i \right | = 7
z-5+8i
Let z be a complex number such that:
z=a+ib
Then the magnitude of z is:
\left| z \right|=\sqrt{a^2+b^2}
Therefore, the magnitude of -5+8i equals:
\sqrt {\left(-5\right) ^ 2 + 8 ^ {2}} = \sqrt {25+64}=\sqrt{89}
\left | -5+8i \right | = \sqrt{89}
z=1-i
Let z be a complex number such that:
z=a+ib
Then the magnitude of z is:
\left| z \right|=\sqrt{a^2+b^2}
Therefore, the magnitude of 1-i equals:
\sqrt {1 ^ 2 + \left(-1\right) ^ {2}} = \sqrt {2}
\left | 1-i \right | = \sqrt{2}