Question

express in the form of a + ib 5+√2i / 1-√2i​

Answer 1

Topic :-

Complex numbers

Given :-

Complex number is $\dfrac{5+i\sqrt2}{1-i\sqrt2}$

To find :-

Representation of given complex number in general form of complex numbers.

Solution :-

We will solve this problem by the concept of conjugate. We will multiply both numerator and denominator with the conjugate of denominator and eliminate iota from the denominator.

Conjugate of a complex number is defined as equal to the complex number but change in the sign of coefficient of iota for the general form of complex number.

Let's say a complex number be ( a + ib ), it's conjugate will be ( a - ib ).

Complex number is denoted by an arrow over the complex number.

Let the denominator of given complex number be $z$.

• $z = 1 - i \sqrt2$

It's conjugate is given by :

• $\overline{z} = 1 + i \sqrt2$

Now multiply both numerator and denominator of given complex number with conjugate.

$\implies \dfrac{5+i\sqrt2}{1-i \sqrt{2} } \times \dfrac{1 + i \sqrt{2} }{1 + i \sqrt{2} }$

‎ ‎ ‎

Apply algebraic identity in denominator :-

• ( A + B ) ( A - B ) = A² - B²

‎ ‎ ‎

$\implies \dfrac{(5+i\sqrt2)(1 + i \sqrt{2})}{(1)^{2} -(i \sqrt{2})^{2} }$

$\implies \dfrac{5+5 \sqrt{2}i + \sqrt{2}i + 2i^{2} }{1-2(i )^{2} }$

‎ ‎ ‎

We know that,

• $i = \sqrt{-1}$
• $i^2 = -1$

‎ ‎ ‎

$\implies \dfrac{5+7 \sqrt{2}i + 2( - 1) }{1-2( - 1 ) }$

$\implies \dfrac{5+7 \sqrt{2}i - 2 }{1 + 2 }$

$\implies \dfrac{3+7 \sqrt{2}i }{3 }$

$\implies \dfrac{3}{3} + \dfrac{7 \sqrt{2}i }{3 }$

$\implies 1+ \dfrac{7 \sqrt{2}i }{3 }$

Hence we have expressed given complex number in its general form a + ib where,

• a = 1
• b = (7 √2) / 3

Know More :-

We need this term iota ( i ) to solve some equations which doesn't have any real roots.

For example:- equation x² + 1 = 0 doesn't have any real roots, we had to either leave these type of equations unsolved if this concept of complex number would be undiscovered.

Every real number can be expressed in the form of complex number a + ib where b = 0.

For example :- Real number say 6 can be expressed as 6 + 0i which is also equals to 1.

iota raise to the power 4k where k is natural number is always equals to 1. In simple words iota raise to the power multiple of 4 is always equals to 1.

Also remember that Set of complex number denoted by C is the universal set for all the numbers present in the universe.

Rational, irrational, whole numbers, real numbers they all belongs to the set of complex numbers. Set of complex numbers is the superset for all these.

Answer 2

Topic :-

Complex numbers

Given :-

Complex number is \dfrac{5+i\sqrt2}{1-i\sqrt2}1−i25+i2

To find :-

Representation of given complex number in general form of complex numbers.

Solution :-

We will solve this problem by the concept of conjugate. We will multiply both numerator and denominator with the conjugate of denominator and eliminate iota from the denominator.

Conjugate of a complex number is defined as equal to the complex number but change in the sign of coefficient of iota for the general form of complex number.

Let's say a complex number be ( a + ib ), it's conjugate will be ( a - ib ).

Complex number is denoted by an arrow over the complex number.

Let the denominator of given complex number be zz .

z = 1 - i \sqrt2z=1−i2

It's conjugate is given by :

\overline{z} = 1 + i \sqrt2z=1+i2

Now multiply both numerator and denominator of given complex number with conjugate.

\implies \dfrac{5+i\sqrt2}{1-i \sqrt{2} } \times \dfrac{1 + i \sqrt{2} }{1 + i \sqrt{2} }⟹1−i25+i2×1+i21+i2

‎ ‎ ‎

Apply algebraic identity in denominator :-

( A + B ) ( A - B ) = A² - B²

‎ ‎ ‎

\implies \dfrac{(5+i\sqrt2)(1 + i \sqrt{2})}{(1)^{2} -(i \sqrt{2})^{2} }⟹(1)2−(i2)2(5+i2)(1+i2)

\implies \dfrac{5+5 \sqrt{2}i + \sqrt{2}i + 2i^{2} }{1-2(i )^{2} }⟹1−2(i)25+52i+2i+2i2

‎ ‎ ‎

We know that,

i = \sqrt{-1}i=−1

i^2 = -1i2=−1

‎ ‎ ‎

\implies \dfrac{5+7 \sqrt{2}i + 2( - 1) }{1-2( - 1 ) }⟹1−2(−1)5+72i+2(−1)

\implies \dfrac{5+7 \sqrt{2}i - 2 }{1 + 2 }⟹1+25+72i−2

\implies \dfrac{3+7 \sqrt{2}i }{3 }⟹33+72i

\implies \dfrac{3}{3} + \dfrac{7 \sqrt{2}i }{3 }⟹33+372i

\implies 1+ \dfrac{7 \sqrt{2}i }{3 }⟹1+372i

Hence we have expressed given complex number in its general form a + ib where,

a = 1

b = (7 √2) / 3

Know More :-

We need this term iota ( i ) to solve some equations which doesn't have any real roots.

For example:- equation x² + 1 = 0 doesn't have any real roots, we had to either leave these type of equations unsolved if this concept of complex number would be undiscovered.

Every real number can be expressed in the form of complex number a + ib where b = 0.

For example :- Real number say 6 can be expressed as 6 + 0i which is also equals to 1.

iota raise to the power 4k where k is natural number is always equals to 1. In simple words iota raise to the power multiple of 4 is always equals to 1.

Also remember that Set of complex number denoted by C is the universal set for all the numbers present in the universe.

Rational, irrational, whole numbers, real numbers they all belongs to the set of complex numbers. Set of complex numbers is the superset for all these.