Mathematics: Find the square roots of each complex number. a. 5 + 12i b. 5 − 12i

Find the square roots of each complex number. a.

Ответ:

Jerryholloway5871

08.01.2022

Mathematics

ANSWER:

Square root of 5 + 12i is 3+2i or -3 – 2i and square root of 5 – 12i is 3 – 2i or -3 + 2i.

SOLUTION:

Given, we have to find square roots of 5 + 12i and 5 – 12i

a) 5 + 12i

Now, square root of 5 + 12i

Suppose that a + bi is a square root of 5 + 12i.

$\text { Then, }(a+b i)^{2}=\left(a^{2}-b^{2}\right)+(2 a b) i=5+12 i$

Equate real and imaginary parts:

$\begin{array}{l}{a^{2}-b^{2}=5 \text { and } 2 a b=12 \rightarrow b=\frac{6}{a}} \\\\ {\text { So, } a^{2}-\left(\frac{6}{a}\right)^{2}=5} \\\\ {\rightarrow a^{2}-\frac{36}{a^{2}}=5} \\\\ {\rightarrow a^{4}-5 a^{2}-36=0} \\\\ {\rightarrow\left(a^{2}-9\right)\left(a^{2}+4\right)=0}\end{array}$

Since a must be real, a = 3 or -3.

This gives b = 2 or -2, respectively.

Thus, we have two square roots: 3+2i or -3-2i.

b) 5 - 12i

Now, square root of 5 - 12i

Suppose that a - bi is a square root of 5 - 12i.

$\text { Then, }(a-b i)^{2}=\left(a^{2}-b^{2}\right)-(2 a b) i=5-12 i$

Equate real and imaginary parts:

$\begin{array}{l}{a^{2}-b^{2}=5 \text { and } 2 a b=12 \rightarrow b=\frac{6}{a}} \\\\ {S o, a^{2}-\left(\frac{6}{a}\right)^{2}=5} \\\\ {\rightarrow a^{2}-\frac{36}{a^{2}}=5} \\\\ {\rightarrow a^{4}-5 a^{2}-36=0} \\\\ {\rightarrow\left(a^{2}-9\right)\left(a^{2}+4\right)=0} \\\\ {\text { since a must be real, } a=3 \text { or }-3}\end{array}$

This gives b = 2 or -2, respectively.

Thus, we have two square roots: 3 - 2i or -3 + 2i.

Hence, square root of 5 + 12i is 3+2i or -3 – 2i and square root of 5 – 12i is 3 – 2i or -3 + 2i.

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Ответ:

juan3937

12.03.2023

Mathematics

Step-by-step explanation:

1,6,9,10,10,11,13,15,15,15

moda-10 and 15

median-10+11/2=10,5

mean-(1+6+9+10+10+11+13+15+15+15)÷10=105÷10=10,5

hope it helps

have a nice day

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Find the square roots of each complex number.
a. 5 + 12i
b. 5 − 12i

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