MHT CET · Maths · Complex Number
If \(Z_1=4 \mathrm{i}^{40}-5 \mathrm{i}^{35}+6 \mathrm{i}^{17}+2, \mathrm{Z}_2=-1+\mathrm{i}\), where \(\mathrm{i}=\sqrt{-1}\), then \(\left|\mathrm{Z}_1+\mathrm{Z}_2\right|=\)
- A 5
- B 13
- C 12
- D 15
Answer & Solution
Correct Answer
(B) 13
Step-by-step Solution
Detailed explanation
\(\mathrm{Z}_1=4 \mathrm{i}^{40}-5 \mathrm{i}^{35}+6 \mathrm{i}^{17}+2\)
\(\mathrm{Z}_1=6+11 \mathrm{i}\)
\(\mathrm{Z}_2=-1+\mathrm{i}\)
\(\therefore \mathrm{Z}_1\) \(+~\mathrm{Z}_2=6+11 \mathrm{i}-1+\mathrm{i}\)
\(=5+12 \mathrm{i}\)
\(\therefore \left|\mathrm{Z}_1+\mathrm{Z}_2\right|=\sqrt{(5)^2+(12)^2}=\sqrt{169}=13\)
\(\mathrm{Z}_1=6+11 \mathrm{i}\)
\(\mathrm{Z}_2=-1+\mathrm{i}\)
\(\therefore \mathrm{Z}_1\) \(+~\mathrm{Z}_2=6+11 \mathrm{i}-1+\mathrm{i}\)
\(=5+12 \mathrm{i}\)
\(\therefore \left|\mathrm{Z}_1+\mathrm{Z}_2\right|=\sqrt{(5)^2+(12)^2}=\sqrt{169}=13\)
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