which complex number has an absolute value of 5?\n-3 + 4i\n2 + 3i\n7 - 2i\n9 + 4i

Answer:

A. $-3 + 4i$

Explanation:

Step1: Recall absolute - value formula

For a complex number $z=a + bi$, $|z|=\sqrt{a^{2}+b^{2}}$.

Step2: Calculate for $z=-3 + 4i$

$| - 3+4i|=\sqrt{(-3)^{2}+4^{2}}=\sqrt{9 + 16}=\sqrt{25}=5$.

Step3: Calculate for $z = 2+3i$

$|2 + 3i|=\sqrt{2^{2}+3^{2}}=\sqrt{4 + 9}=\sqrt{13}\neq5$.

Step4: Calculate for $z = 7-2i$

$|7 - 2i|=\sqrt{7^{2}+(-2)^{2}}=\sqrt{49+4}=\sqrt{53}\neq5$.

Step5: Calculate for $z = 9+4i$

$|9 + 4i|=\sqrt{9^{2}+4^{2}}=\sqrt{81 + 16}=\sqrt{97}\neq5$.