(-√5 +2√-4)+(1-√-9)+(2+3i)(2-3i) then find value of a and b
Answers
Answered by
69
Answer:
Step-by-step explanation:
Given:
separating real and imaginary parts, we get
Answered by
0
Answer:
Answer:
a=14-\sqrt5a=14−
5
b=1b=1
Step-by-step explanation:
Given:
(-\sqrt5 +2\sqrt{-4})+(1-\sqrt{-9})+(2+3i)(2-3i)=a+ib(−
5
+2
−4
)+(1−
−9
)+(2+3i)(2−3i)=a+ib
(-\sqrt5 +2\sqrt{4(-1)})+(1-\sqrt{9(-1)})+2^2-(3i)^2=a+ib(−
5
+2
4(−1)
)+(1−
9(−1)
)+2
2
−(3i)
2
=a+ib
(-\sqrt5 +2\sqrt{2^2i^2})+(1-\sqrt{3^2i^2})+2^2-3^2i^2=a+ib(−
5
+2
2
2
i
2
)+(1−
3
2
i
2
)+2
2
−3
2
i
2
=a+ib
(-\sqrt5 +2(2i))+(1-3i)+4-9(-1)=a+ib(−
5
+2(2i))+(1−3i)+4−9(−1)=a+ib
(-\sqrt5 +4i)+(1-3i)+4+9=a+ib(−
5
+4i)+(1−3i)+4+9=a+ib
(-\sqrt5 +4i)+14-3i=a+ib(−
5
+4i)+14−3i=a+ib
\implies\:(14-\sqrt5)+i=a+ib⟹(14−
5
)+i=a+ib
separating real and imaginary parts, we get
a=14-\sqrt5a=14−
5
b=1b=1
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