Asked by Kayla Valdes on May 12, 2024
Verified
Find the multiplicative inverse of the complex number 5+6i5 + 6 i5+6i .
A) −561+661i- \frac { 5 } { 61 } + \frac { 6 } { 61 } i−615+616i
B) 561+661i\frac { 5 } { 61 } + \frac { 6 } { 61 } i615+616i
C) −561−661i- \frac { 5 } { 61 } - \frac { 6 } { 61 } i−615−616i
D) 561−661i\frac { 5 } { 61 } - \frac { 6 } { 61 } i615−616i
E) 5+6i5 + 6 i5+6i
Multiplicative Inverse
The number that, when multiplied by a given number, results in a product of one. It is also known as the reciprocal.
Complex Number
A value represented as a + bi, where a and b are real numbers, and i is the unit of imaginary numbers.
- Recognize and utilize the principles of additive inverse and multiplicative inverse within the framework of complex numbers.
Verified Answer
MA
Maidul AhsanMay 17, 2024
Final Answer :
D
Explanation :
The multiplicative inverse of a complex number is found by dividing 1 by the complex number. Therefore, we need to solve the equation:
(5+6i)(a+bi)=1(5+6i)(a+bi)=1(5+6i)(a+bi)=1
where $a+bi$ is the multiplicative inverse of $5+6i$. Expanding the left side of the equation gives:
5a+6ai+5bi+6bi2=15a+6ai+5bi+6bi^2=15a+6ai+5bi+6bi2=1
Since $i^2=-1$, we can simplify to:
(5a−6b)+(6a+5b)i=1(5a-6b)+(6a+5b)i=1(5a−6b)+(6a+5b)i=1
Now we have two equations that must be satisfied simultaneously:
5a−6b=15a-6b=15a−6b=1
6a+5b=06a+5b=06a+5b=0
Solving the second equation for $b$ gives $b=-\frac{6a}{5}$, which can be substituted into the first equation:
5a−6(−6a5)=15a-6\left(-\frac{6a}{5}\right)=15a−6(−56a)=1
Simplifying yields:
a=561a=\frac{5}{61}a=615
b=−661b=-\frac{6}{61}b=−616
Therefore, the multiplicative inverse of $5+6i$ is $\boxed{\textbf{(D)}\ \frac{5}{61}-\frac{6}{61}i}$.
(5+6i)(a+bi)=1(5+6i)(a+bi)=1(5+6i)(a+bi)=1
where $a+bi$ is the multiplicative inverse of $5+6i$. Expanding the left side of the equation gives:
5a+6ai+5bi+6bi2=15a+6ai+5bi+6bi^2=15a+6ai+5bi+6bi2=1
Since $i^2=-1$, we can simplify to:
(5a−6b)+(6a+5b)i=1(5a-6b)+(6a+5b)i=1(5a−6b)+(6a+5b)i=1
Now we have two equations that must be satisfied simultaneously:
5a−6b=15a-6b=15a−6b=1
6a+5b=06a+5b=06a+5b=0
Solving the second equation for $b$ gives $b=-\frac{6a}{5}$, which can be substituted into the first equation:
5a−6(−6a5)=15a-6\left(-\frac{6a}{5}\right)=15a−6(−56a)=1
Simplifying yields:
a=561a=\frac{5}{61}a=615
b=−661b=-\frac{6}{61}b=−616
Therefore, the multiplicative inverse of $5+6i$ is $\boxed{\textbf{(D)}\ \frac{5}{61}-\frac{6}{61}i}$.
Learning Objectives
- Recognize and utilize the principles of additive inverse and multiplicative inverse within the framework of complex numbers.