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Question

Reduce (114i21+i)(34i5+i) to the standard form.

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Solution

Given: (114i21+i)(34i5+i)
=[(1+i)2(14i)(14i)(1+i)](34i5+i)
=(1+i2+8i1+i4i4i2)(34i5+i)
=(1+9i53i)(34i5+i)
=(3+4i+27i36i225+5i15i3i2)
=33+31i2810i=33+31i2(145i)
=(33+31i)2(145i)×(14+5i)(14+5i) [ on multiplying numerator and denominator by (14+5i) ]
=462+165i+434i+155i22[(14)2(5i)2]=307+599i2(19625i2)
=307+599i2(221)=307+599i442=307442+599i442

which is the required standard form.

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