Find the real and imaginary parts of z when: 1/z = (2/(2+3i)) + (1/(3-2i))

Question

The answer on the book is 7/5 + i 4/5. However, I couldn't solve it.
Please show step by step.

Answer 1

Hello Suzy,

Here are the steps I took:

i) Found a common denominator on the right side to add the fractions

ii) Took the reciprocals of both sides

iii) Multiplied top and bottom by the conjugate to get 'i' out of the bottom.

 

Given problem: 1/z = (2/(2+3i)) + (1/(3-2i))

 

i) 1/z = (2/(2+3i))*[(3-2i)/(3-2i)] + (1/(3-2i))*[(2+3i)/(2+3i)]

= [(6-4i)+(2+3i)] / (6+5i-6i²)

→ 1/z = (8-i) / (12+5i)  *Note i² = -1

 

ii) z = (12+5i) / (8-i)

 

iii) z = [(12+5i) / (8-i)] * [(8+i)/(8+i)]

= (96+52i+5i²) / (64-i²)

= (91+52i) / 65

= (7+4i) / 5

= 7/5 +(4/5)i

 

Hope the above helps, and thank you for posting the question.  If you have any questions about any of the steps, please let me know.

 

Kind regards,

Michael Ehlers

Answer 2

We find (1/z) first, but will not simplify.

 

1/z = (2/(2+3i)) + (1/(3-2i)).

 

We add the fractions by first finding LCD and adding the denominators.

 

1/z = [2(3-2i) + 1(2+3i)]/[(2+3i)(3-2i)],

 

1/z = [6-4i+2+3i]/[6-4i+9i -6i²],

 

1/z = (8-i)/(12+5i).

 

It is time to find z.

 

z = (12+5i)/(8-i).

 

You can rationalize the denominator by multiplying by the complex conjugate (8+i) and simplify.

 

Any questions?