# Find the answer to the following problem (1.5 + 3.2i) – (-2.4 – 3.7i)

• Last Updated : 21 Mar, 2022

Complex number is the term that can be shown as a sum of a real number and an imaginary number. A complex number can be written in the form of a+ib, where a and b both are real numbers. It is denoted by z.

For example:

6+2i is a complex number, where 3 is a real number (Re) and 4i is an imaginary number (Im).

9+5i is a complex number where 2 is a real number (Re) and 5i is an imaginary number (im)

The real number and imaginary number combination is called a Complex number.

Example: √-5, -8i, -13i are all imaginary numbers. here ‘i’ is an imaginary number called “iota’

### Find the answer to the following problem (1.5 + 3.2i) – (-2.4 – 3.7i)

Given: (1.5 + 3.2i) – (-2.4 – 3.7i)

By open bracket

= 1.5 + 3.2i + 2.4 + 3.7i

= 3.9 + 9.9i

### Similar Questions

Question 1: Simplify  (-i)(2i)(-1/8i)3?

Solution:

Given: (-i)(2i)(-1/8i)3

= (-i)(2i) (-1/8i)3

= -2i2 (-1/512i3)

= -2i (-1/512 i3)

= -2(-1) {(-1/512)(- i)}                {i2 = -1 and i3 = -i}

= 2(1/512i)

= (2/512)i

= 0 + 1/256i

Question 2: Express the (1-i) – (-1+6i)?

Solution:

Given : (1-i) – (-1+6i)

= 1 – i + 1 – 6i

= 2 – 7i

Question 3: Simplify ( 5 -3i)3?

Solution:

Given: (5 -3i)3

Here we will use identity (a-b)3 = a3 – b3 – 3a2b + 3ab2

= (5)3 – (3i)3 – 3(5)2(3i) + 3 (5) (3i)2

= 125 – (27i3) – 225i + 15(9i2)

= 125 -[27(-i)] – 225 i + 15(-9)

= 125 +27i -225i – 135

= -10 – 198i

Question 4: Express the {1/5 + 2/5i} – {4 + 5/2i}

Solution:

Given: {1/5 +2/5 i} – { 4 + 5/2 i}

= {(1+2i)/5} – {(8+5i)/ 2}

= [{2(1+2i) – 5(8+5i)} /10]

= {2+4i-40 -25i} /10

= (-38 -21i)/10

= -38/10 – 21/10 i

= -19/5 – 21/10 i

Question 5: Simplify (-6i)(7i)(-2)

Solution:

Given: (-6i)(7i)(-2)

= -6i x 7i x (-2)

= -42i2 x -2                            {i2 = -1}

= -42 (-1) x -2

= 42 x -2

= -84 + 0i

Question 6: Prove that the {(2+3i) / (3+4i)} {(2-3i)/(3-4i)} complex numbers are purely real?

Given : {(2+3i) / (3+4i) } { (2-3i)/(3-4i) }

= {(2+3i)(2-3i) } / {(3+4i)(3-4i)}

= {4 -6i +6i -9(i2)} / {9 -12i + 12i – 16(i2)}

= {4 +9} / {9 +16}

= 13/25 + 0i

Therefore {(2+3i) / (3+4i)} {(2-3i)/(3-4i)} it is purely real

Question 7: Perform the indicated operation and write the answer in standard form (3-i)/(1+2i)?

Solution:

Given : (3-i)/(1+2i)

Multiply with the conjugate of denominator

= {(3-i)/(1+2i) x (1-2i)/(1-2i)}

= {(3-i)(1-2i)} / {(1)2 -(2i)2}                {difference of squares formula . i.e (a+b)(a-b) = a2 – b2}

= {3 -6i -i + 2i2} / {1-4(-1)}

= {3-7i -2} / {1 + 4}                        { i2 = -1 }

= (1-7i)/5

= 1/5 -7/5 i

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