Given that,
Voltage = 34 volt
Current = 3i mA
We need to calculate the complex number to represent the impedance
Using ohm's law
![V=IR](https://img.qammunity.org/2021/formulas/physics/high-school/8gv7kiyicaqmg4zdl8vw77xnncjzk66hn7.png)
![R=(V)/(I)](https://img.qammunity.org/2021/formulas/mathematics/high-school/v9x3ncwxjjvncmudk2jt6mycefnhd63qdg.png)
Where, V = voltage
I = current
R = impedance
Put the value into the formula
![R=(34)/(0+0.003i)*(0-0.003i)/(0-0.003i)](https://img.qammunity.org/2021/formulas/mathematics/high-school/5zp5kodymtdk0u2pfbdpatv20g5y9mc97z.png)
![R=(0.102i)/(0.9*10^(-5))\ \Omega](https://img.qammunity.org/2021/formulas/mathematics/high-school/btm6lja9obetqhn54cpiir3hnqcwv15uua.png)
(b). Given that,
Voltage = 13 volts
Current = 2.4 mA
We need to calculate the complex number to represent the impedance
Using ohm's law
![R=(V)/(I)](https://img.qammunity.org/2021/formulas/mathematics/high-school/v9x3ncwxjjvncmudk2jt6mycefnhd63qdg.png)
Put the value into the formula
![R=(13)/(0.00024i)*(-0.00024i)/(-0.00024i)](https://img.qammunity.org/2021/formulas/mathematics/high-school/efqdo8274s5l5p16jw64uvn9lf3602d7ly.png)
![R=(0.00312i)/(5.76*10^(-8))\ \Omega](https://img.qammunity.org/2021/formulas/mathematics/high-school/t68xr1b1fxmlc4scjx5nzt8q53qyqaxzal.png)
Hence, (a). The complex number to represent the impedance is
![(0.102i)/(0.9*10^(-5))\ \Omega](https://img.qammunity.org/2021/formulas/mathematics/high-school/qdbmk2qkf4ab4tya14i833sgn3dbthm0ie.png)
(b). (a). The complex number to represent the impedance is
![(0.00312i)/(5.76*10^(-8))\ \Omega](https://img.qammunity.org/2021/formulas/mathematics/high-school/hiqg92nr51su24guwms4y1nh8yzu1xoyd6.png)