**How to Show that Given Complex Number is Purely Real and Imaginary :**

Here we are going to see some example problems to understand how to Show that Given Complex Number is Purely Real and Imaginary.

**Question 1 :**

Find the least value of the positive integer n for which (√3 + i)^{n }(i) real (ii) purely imaginary.

**Solution : **

(i) Purely real

If n = 1

= (√3 + i)^{1}

= (√3 + i)

If n = 2

= (√3 + i)^{2}

= √3^{2} + i^{2 }+ i2√3

= 2 + i 2√3

If n = 3

= (√3 + i)^{3}

= √3^{3} + 9i^{ }+ 3√3(-1) - i

= 3√3 + 9i^{ }- 3√3 - i

= 8i (purely imaginary)

If n = 6

= (√3 + i)^{6}

= {(√3 + i)^{3}}^{2}

= (8i)^{2}

= 64 i^{2}

= -64 (purely real)

Hence the least value of n is 6 to make (√3 + i)^{n }as purely real, make it as purely imaginary, we have to put n = 3.

**Question 2 :**

Show that (i) (2 + i√3)^{10} - (2 - i√3)^{10 }is purely imaginary

**Solution :**

**Let z = (2 + i√3) ^{10}, then z bar = **(2 - i√3)

By using properties of conjugate, we get

(z - z bar)/2 = Im (z)

Hence the given is purely imaginary.

(ii) [(19 - 7i)/(9 + i)]^{12} + [(20 - 5i)/(7 - 6i)]^{12 }is real

**Solution :**

Let z = (19 - 7i)/(9 + i)

= [(19 - 7i)/(9 + i)][(9 - i)/(9 - i)]

= [(19 - 7i)(9 - i)/(9 + i)(9 - i)]

= (171 - 19i - 63i + 7i^{2}) / (81 - (1))

= (164 - 82i) / 82

z = 2 - i

(20 - 5i)/(7 - 6i)

= [(20 - 5i)/(7 - 6i)] [(7 + 6i)/(7 + 6i)]

= [(20 - 5i)(7 + 6i)/(7 - 6i) (7 + 6i)]

= (140 + 120i - 35i - 30i^{2})/(49 - 36(-1))

= (140 + 120i - 35i + 30)/(49 + 36)

= (170 + 85i)/85

z bar = 2 + i

By adding a complex number and its conjugate, we get the real part of the complex number.

Hence the sum of above complex numbers is purely real.

After having gone through the stuff given above, we hope that the students would have understood, "How to Show that Given Complex Number is Purely Real and Imaginary".

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