Question 5, Exercise 1.3
Solutions of Question 5 of Exercise 1.3 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.
Question 5(i)
Find the solutions of the equation z2+z+3=0.
Solution
Given:
z2+z+3=0.
According to the quadratic formula, we have
a=1, b=1 and c=3.
Thus we have
z=−b±√b2−4ac2a=−1±√(1)2−4(1)(3)2(1)=−1±√1−122=−1±√112i
Thus the solutions of the given equation are −12±√112i.
Question 5(ii)
Find the solutions of the equation z2−1=z.
Solution
Given: z2−1=z
$$\implies {{z}^{2}}-z-1=0$$
According to the quadratic formula, we have
a=1, b=−1 and c=−1
So we have
z=−b±√b2−4ac2a=−(−1)±√(−1)2−4(1)(−1)2(1)=1±√1+42=1±√52.
Thus the solutions of the given equations are 1±√52.
Question 5(iii)
Find the solutions of the equation z2−2z+i=0
Solution
Given: z2−2z+i=0
According to the quadratic formula, we have
a=1, b=−2 and c=i
So we have
z=−b±√b2−4ac2a=−(−2)±√(−2)2−4(1)(i)2(1)=2±√4−4i2=2±2√1−i2=1±√1−i
Thus the solutions of the given equations are 1±√1−i.
Question 5(iv)
Find the solutions of the equation z2+4=0
Solution
Given: z2+4=0 \begin{align} \implies &z^2-(2i)^2=0\\ \implies &(z+2i)(z-2i)=0\\ \implies &z+2i=0 \quad \text{or} \quad z-2i=0\\ \implies &z=-2i \quad \text{or} \quad z=2i.\end{align} Thus, the solutions of the given equations are ±2i.
Go To