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Question: The modulus of \[\left( {1 + i} \right)\left( {3 + 4i} \right) = \] A. \[\sqrt {50} \] B. \[\sqr...

The modulus of (1+i)(3+4i)=\left( {1 + i} \right)\left( {3 + 4i} \right) =
A. 50\sqrt {50}
B. 25\sqrt {25}
C. 10
D. 10210\sqrt 2

Explanation

Solution

GComplex numbers are the numbers that are expressed in the form of a+iba + ib where, a, b are real numbers and ‘i’ is an imaginary number. Here in the given question, we have the terms as; (1+i)(3+4i)\left( {1 + i} \right)\left( {3 + 4i} \right), in which 1 and 3 are real numbers and i, 4i are complex numbers, hence here to find the modulus of the given complex numbers we need to just multiply by applying modulus to both the terms.

Formula used:
S∞=a1−r{S_\infty } = \dfrac{a}{{1 - r}}
S∞{S_\infty } = Sum of infinite geometric progression
a = First term
r = Common ratio
n = Number of terms

Complete step by step solution:
(1+i)(3+4i)\left( {1 + i} \right)\left( {3 + 4i} \right) ………………. 1
z1,z2{z_1},{z_2} being two complex numbers, then we have:
∣z1⋅z2∣=∣z1∣⋅∣z2∣\left| {{z_1} \cdot {z_2}} \right| = \left| {{z_1}} \right| \cdot \left| {{z_2}} \right| ………………. 2
According to the given complex numbers, from equation 1 we have equation 2 as:
∣z1⋅z2∣=∣z1∣⋅∣z2∣\left| {{z_1} \cdot {z_2}} \right| = \left| {{z_1}} \right| \cdot \left| {{z_2}} \right|
⇒∣(1+i)(3+4i)∣=∣1+i∣⋅∣3+4i∣\Rightarrow \left| {\left( {1 + i} \right)\left( {3 + 4i} \right)} \right| = \left| {1 + i} \right| \cdot \left| {3 + 4i} \right|
To evaluate the terms, we need to simplify it by squaring the terms as:
=12+12×32+42= \sqrt {{1^2} + {1^2}} \times \sqrt {{3^2} + {4^2}}
Evaluating the terms, we get:
=2×9+16= \sqrt 2 \times \sqrt {9 + 16}
=2×25= \sqrt 2 \times \sqrt {25}
Simplify the terms, we have:
=50= \sqrt {50}
Therefore, we have:
(1+i)(3+4i)=50\left( {1 + i} \right)\left( {3 + 4i} \right) = \sqrt {50}

So, the correct answer is Option A.

Additional information:
Any number which is present in a number system such as positive, negative, zero, integer, rational, irrational, fractions, etc. are real numbers. The numbers which are not real are imaginary numbers. When we square an imaginary number, it gives a negative result are all imaginary numbers.

Note: We must know that the combination of both the real number and imaginary number is a complex number. And we must note the imaginary number is usually represented by ‘i’ or ‘j’, which is equal to −1\sqrt { - 1} . Therefore, the square of the imaginary number gives a negative value i.e., the values of i2=−1{i^2} = - 1 and i=−1i = \sqrt { - 1} . As we know, 0 is a real number and real numbers are part of complex numbers, therefore, 0 is also a complex number and is represented as 0+0i0 + 0i.