Question

Find the conjugate and modulus of the complex number
3+2i / 2-5i + 3-2i / 2+5i
​

Answer 1

Answer:

$\frac{3 + 2i}{2 - 5i} + \frac{3 - 2i}{2 + 5i}$

$\frac{(3 + 2i)(2 + 5i) + (2 - 5i)(3 - 2i)}{ {2}^{2} + 5^{2} }$

$\frac{6 + 19i - 10i + 6 - 19i - 10i}{29}$

$\frac{12 - 20i}{29}$

Conjugate

$\frac{12 + 20i}{29}$

Modulus

$\sqrt{ \frac{12 {}^{2} + 20 {}^{2} }{29 {}^{2} } }$

$= \frac{1}{29} \sqrt{ {544} }$

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Answer 2

Answer: Conjugate = $Z'=\frac{3+2i}{2-5i} -\frac{3-2i}{2+5i}$ and modulus =$\sqrt{\frac{64}{841} }$.

Step-by-step explanation:

Concept: A complex number is represented as  the sum of a real number and an imaginary number. It is of the form a + ib where a and b are both real numbers and i stands for iota which takes the value as $\sqrt{-1}$ which is the imaginary part and hence a forms the real part and b*i forms the imaginary part.

Modulus of a complex number: The complex number's modulus is the the distance of the point on the argand plane representing the complex number Z from the origin. P shall stand for the point designating the complex number Z = x + iy. Then, OP = |Z| = $\sqrt{x^{2} +y^{2} }$ .

Conjugate of a complex number: It is obtained by changing the sign between the real and the imaginary part of the complex number.

Step 1: let,  $Z=\frac{3+2i}{2-5i} +\frac{3-2i}{2+5i}$

Now,

$Z=\frac{3+2i}{2-5i} +\frac{3-2i}{2+5i} \\=\frac{(3+2i)(2+5i)+(3-2i)(2-5i)}{(2-5i)(2+5i)} \\= \frac{6+15i+4i+10i^{2}+6-15i-4i+10i^{2} }{(4-25i^{2}) } \\= - \frac{8}{29}+i*0$

= a+bi, where a= $-\frac{8}{29}$ and b = 0

Step 2: Now in order to find the modulus of the Z, we follow the steps:

|Z| =$\sqrt{a^{2}+b^{2} }$= $\sqrt{(-\frac{8}{29})+0^{2} }$

                    = $\sqrt{\frac{64}{841} }$ = modulus of Z

Step 3: Now to get the conjugate of the complex number, we proceed as follows:

$Z=\frac{3+2i}{2-5i} +\frac{3-2i}{2+5i}$

⇒ $Z'=\frac{3+2i}{2-5i} -\frac{3-2i}{2+5i}$

So here Z' is the conjugate of the complex number Z.