Find the square root of (-16)
Complex numbers are terms that can be shown as the sum of real and imaginary numbers. These are the numbers that can be written in the form of a + ib, where a and b both are real numbers. It is denoted by z. Here the value βaβ is called the real part which is denoted by Re(z), and βbβ is called the imaginary part Im(z) in form of a complex number. It is also called an imaginary number. In complex number form a + bi βiβ is an imaginary number called βiotaβ. The value of i is (β-1) or we can write as i2 = -1. For example,
3 + 16i is a complex number, where 9 is a real number (Re) and 16i is an imaginary number (Im).15 + 20i is a complex number where 10 is a real number (Re) and 20i is an imaginary number (Im)
Find the square root of (-16)
Solution:
As we have discussed above about the complex number
Now as per the question to find the square root of (-16).
= β-16
= β16(-1)
= (β16)(β-1) {i = β-1}
= 4i
Similar Questions
Question 1: Find the value of square root of {-25}?
Solution:
Given: β-25
= β-25
= β25(-1)
= (β25)(β-1) {i = β-1}
= 5i
Question 2: Find the value of square root of {-289}?
Solution:
Given : β-289
= β-289
= β289(-1)
= (β289)(β-1) {i = β-1}
= 17i
Question 3: Find the square of (9i)?
Solution:
After squaring an imaginary number, it gives a result in negative β¦
(9i)2 = 9i Γ 9i
= 81i2
= 81(-1)
= -100
Question 4: Simplify {(-3 β 5i) / (3 +2i)}?
Solution:
Given {(-3 β 5i) / (3 +2i)}
Conjugate of denominator 3+2i is 3-2i
Multiply with the conjugate of denominator
Therefore {(-3 β 5i) / (3 +2i)} x {(3-2i) / (3-2i)}
= {(-3-5i)(3-2i)} / {(3+2i)(3-2i)}
= {-9 +6i -15i +10i2 } / {32 β (2i)2} {difference of squares formula . i.e (a+b)(a-b) = a2 β b2 }
= {-9 + 6i -15i + 10 (-1)} / {9 β 4(-1)} { i2 = -1 }
= {-9 + 6i -15i -10} / {9 + 4}
= (-19 β 9i) / 13
= -19 /13 β 9i /13
= -19/13 β 9/13 i
Question 5: Simplify the given expression 7/10i
Solution:
Given: 7/10i
Standard form of numerator, 7 = 7 +0i
Standard form of denominator, 10i = 0 +10i
Conjugate of denominator, 0 + 10i = 0 β 10i
Multiply the numerator and denominator with the conjugate,
Therefore, {(7 + 0i) / (0 + 10i)} Γ {(0 β 10i)/(0 β 10i)}
= {7(0 β 10i)} / {0 β (10i)2}
= {0 β 70i} / {0 β (100(-1))}
= {-70i} / 100
= 0 β 70/100i
= -7/10i
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